Soumen Mondal, Author at Labkafe Blog

Spherometer – Measure the radius of curvature of a spherical surface | Labkafe

Soumen Mondal — Fri, 17 Mar 2023 07:57:06 +0000

Aim:

To determine the radius of curvature of a given spherical surface by a Spherometer.

Apparatus:

Spherometer
Glass Slab
Half Meter Scale
Convex Lens

Theory:

A spherometer is a measuring instrument used to measure the radius of curvature of a spherical surface and a very small thickness.

Figure 3.1 is a schematic diagram of a single disk spherometer. It consists of a central leg OS, which can be raised or lowered through a threaded hole V (nut) at the centre of the frame F. The metallic triangular frame F supported on three legs of equal length A, B and C. The lower tips of the legs form three corners of an equilateral triangle ABC and lie on the periphery of a base circle of known radius, r. The lower tip of the central screw, when lowered to the plane (formed by the tips of legs A, B and C) touches the centre of triangle ABC. A circular scale (disc) D is attached to the screw. The circular scale may have 50 or 100 divisions engraved on it. A vertical scale P marked in millimetres or half-millimetres, called main scale or pitch scale P is also fixed parallel to the central screw, at one end of the frame F. This scale is kept very close to the rim of disc D but it does not touch the disc D. This scale reads the vertical distance when the central leg moves through the hole V.

Fig 3.1

Principle:

Pitch of a Spherometer

The vertical distance moved by the screw S in one complete rotation of the circular Scale/Disc D is called the pitch (p) of the spherometer. To find the pitch, give full rotation to the screw (say 4 times) and note the distance (d) advanced over the pitch scale.

If the distance d is 4 mm The pitch can be represented as,

Least Count of the Screw Gauge

The Least count (LC) is the distance moved by the spherometer screw, when the screw is turned through 1 division on the circular. We are using a spherometer which has 100 divisions (N) on the disc. The least count can be calculated using the formula,

The formula for the radius of curvature of a spherical surface

Approach 1:

From the figure 3.3, O is the centre of the circle. OE = OA = R, radius of the circle. F is the tip of the screw at the same plane with A, B and C. EF = h, AF = a and ∠AFO =

900.

Therefore, geometrically we can write,

OA2 = OF2 + FA2

or, R2 = (R-h)2 + a2

= R2 -2.R.h + h2 + a2

∴ R = (h2 + a2 )/2h

Now, let l be the distance between any two legs of the spherometer as shown in figure 3.6, then from geometry we have, a = . Thus the radius of curvature of the spherical surface can be given by,

∴ R = ( 3h2 + l2 )/6h

Fig 3.3

Approach 2:

From the figure 3.4, the circle is passing through A and C. O is the centre of the circle. OE =R, radius of the circle. F is the tip of the screw at the same plane with A, B and C. CF = h, AF = a ∠EAC = 900.

∴ CE2 = AE2 + AC2

or, (2R2) = (AF2 + FE2) + (CF2 + AF2)

= a2 + (2R -h)2 + h2 + a2

∴ R= a2/2h+ h/2

Now, let l be the distance between any two legs of the spherometer or the side of the equilateral triangle ABC (Fig. 3.4), then from geometry we have, a = l/√3. Thus the radius of curvature of the spherical surface can be given by,

R = a2/2h+ h/2

or, R = l2/6h+ h/2

Diagram:

Fig 3.4

Fig 3.5

Fig. 3.6

Procedure:

Find the pitch (p) of the screw and count the total number of divisions (N) in the circular scale.
Place the spherometer in the plane glass plate. Now rotate the head T anti-clockwise to raise the tip of the central screw S by a certain distance.
Place the spherometer on the convex surface. Gently rotate T clockwise to bring down the tip of S until it just touches the spherical surface. Use a paper strip and try to pass between the tip of the screw and spherical surface to check if there is no gap between them.
Record the initial circular scale reading (r1) in table 3.1. Circular scale reading means the divisions engraved on the disc which coincides with the linear scale.
Place the spherometer on the glass slab without disturbing the initial circular scale reading (c.s.r). Then slowly rotate T clockwise to bring the tip down and touch the glass plate. During this rotation count the number of full rotation (n) of the circular scale. Take the final c.s.r. (r2) when the tip touches the glass plate.
Repeat step 2 and 5 at least thrice by placing the spherometer at different places.
Now, place the spherometer on a piece of paper and press it lightly so that an imprint of the three legs is made on the paper. You can do it on your laboratory notebook on the left side white page.
Measure each side of the triangle AB, BC, and CA formed by the points (A, B, C).
Take mean of them. Thus we get l.

Observations:

Least count of spherometer :

Total number of divisions is the in circular scale, N = _______

One linear scale division, L.S.D. = ____ mm

Distance moved by the screw for 4 rotations, d = ________ mm

Pitch of the screw, p = 4/d = ____mm

Therefore, Least Count, L.C. = p/N= ______________mm

Distance between two legs of spherometer:

AB = _______ cm, BC = _______cm, CA = _________cm

∴ l = (AB + BC+ CA)/3 = _______________________cm

Table 3.1 Table for height (h)

Mean value of sagitta, h = _________________ mm = ________________cm

Calculation:

Radius of curvature of the given convex surface, R = (3h2 + l2 )/6h =…………………………..cm

or, Radius of curvature of the given convex surface, R = l2/6h+ h/2 =…………………………… cm

Precautions:

The screw should move freely without friction.
The screw should be rotated in one direction to get any reading. Otherwise back-lash error will be introduced.
The circular should not be rotated any more, even slightly, when it touches a surface.
The linear scale is not used to take readings and h is calculated by taking the difference of two circular scale readings. Hence we do not need to find the zero error of the instrument.

Reference:

http://www.ncert.nic.in/

Your may checkout our blog on SCREW GAUGE & LEAST COUNT

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How to find the least count of vernier calipers and measure the diameter of a body | Labkafe

Soumen Mondal — Sat, 31 Jul 2021 07:42:57 +0000

Aim:
(a) To find the least count of Vernier calipers

(b) To measure the diameter of a small spherical/cylindrical body

(c) To measure internal diameter and depth of given beaker using Vernier Calipers and (d) hence find its volume

Apparatus:

Vernier Calipers
Cylinder (Metal)
Beaker (Graduated)

Theory:

A Vernier Caliper has two scales–one main scale and a Vernier scale. The Vernier Scale slides along the main scale. Generally, each division of the Vernier scale is smaller than each division of main scale. In our example, nine main scale divisions are equal to ten Vernier scale divisions.

The main scale is graduated in cm and mm. It has two fixed jaws, A and C, projected at right angles to the scale. The sliding Vernier scale has jaws (B, D) projected at right angles to it and a metallic strip (N). The zero mark on the main scale and the Vernier scale coincide with each other when the jaws touch. Moreover, the jaws (A, B) are used to measure the external diameter of an object while jaws (C, D) are used to measure the internal diameter of an object.

The metallic strip (N) is designed to measure the height / depth of objects. Knob P is used to slide the Vernier scale on the main scale. Screw S is used to fix the Vernier scale at a desired position. The least count of a common meter scale is 0.1 cm or 1 mm. It is difficult to decrease the least count further, as then the graduations would be difficult to see. A Vernier scale enables more precise measurement, as here two different scales are used for measurement.

Fig. 1.1

Determination of Least Count of Vernier Calipers or Vernier Constant:

The difference in the magnitude of one main scale division (M.S.D.) and one Vernier scale division (V.S.D.) is called the least count of vernier Calipers. This is the smallest distance that can be measured using the instrument. The Least Count (LC) is also called the Vernier Constant (VC).

In Fig. 1.1, 10 divisions on the Vernier scale coincide with 9 divisions on the main scale and the length of 1 division on the main scale is 1 mm.

10 Vernier Scale Divisions (VSD) = 9 Main Scale Divisions (MSD)

1 VSD = 0.9 MSD

The quantity (1 MSD – 1 VSD) is called the Vernier Constant (VC).

VC = 1 MSD – 1 VSD = 1 mm – 0.9 mm

Hence, VC = 0.1mm = 0.01cm

The volume of the spherical object is:

V = (4/3)πr 3, where r is the radius of the sphere.

Determination of Zero Error:

If the zero mark on the main scale coincides with the zero mark on the Vernier scale when jaws C and D are brought in contact with each other, then the instrument is free from zero error as shown in fig. 1.2. However, it is never so. Because of wear and tear of the jaws or due to manufacturing defects, the zero mark on the Vernier scale does not coincide with the zero mark on the main scale. This gives rise to zero error. Zero error can be positive or negative.

Fig. 1.2

(i) Positive zero error and its correction:

The zero error is positive when the zero mark on the Vernier scale shifts towards the right side of the zero mark on the main scale when jaws C and D touch each other. In this case, the measured length will be more than the actual length and, therefore, the zero error is called positive zero error. Hence, we have to subtract the positive error from the obtained reading. If, for example, the 3rd Vernier mark coincides with any main scale mark as shown in Fig 1.3, then the Zero error is:

Zero error = 0.00 + 3 x Least Count = 0.00 + 3 x 0.01 cm

Zero error = 0.03 cm

Hence, the actual reading = Measured reading – (0.03) cm

Fig. 1.3

(ii) Negative zero error and its correction:

The zero error is negative when the zero mark on the Vernier scale shifts towards the left side of the zero mark on the main scale when jaws C and D touch each other. In this case, the measured length will be less than the actual length and, therefore, the zero error is called negative zero error. Hence, we have to add negative error with the obtained reading. If, for example, the 8th Vernier mark coincides with a main scale mark as shown in Fig 1.4, then the Zero error is:

Zero error = 0.00 – (10-8) ** x Least Count = 0.00 – 2 x 0.01

** To calculate the negative zero error, we subtract the mark on the vernier scale that coincides with any mark on the main scale, from 10, before multiplying it with the least count.

Zero error = -0.02 cm

Hence, the actual reading = Measured reading – (-0.02)

Fig. 1.4

Fig. 1.5

Fig. 1.6

Fig. 1.7

Procedure:

Determine the Vernier constant (V.C.) i.e., least count of Vernier calipers as explained above and record it.
Bring the movable jaw BD in close contact with the fixed jaw AC and find the zero error. Do it three times and record it. If there is no zero error, then record zero error as Nil. If it does not coincide, then find the zero error and record it.

Diameter of spherical body

With the help of lower jaws (A, B) grip the spherical body gently without exerting excess pressure, as shown in Fig. 1.5. Tighten the screw S attached to the Vernier scale.
Record the main scale reading (N) just before the zero mark of the Vernier scale in Table 1.1.
Note the number (n) of the Vernier scale division which coincides with any division of the main scale.
Repeat the steps given above after rotating the body by 90°. Measure the diameter in a direction perpendicular to the original direction.
Take at least four observations along different directions and apply zero error correction.
Take the mean of all recorded diameters.

Internal diameter of the beaker

Put the jaws C and D inside the rim of the beaker and open them till each jaw touches the inner wall of the beaker, without any undue pressure on the walls as shown in fig. 1.6. Tighten the screw S attached to the Vernier scale gently.
Note the-position of the zero mark of the Vernier scale on the main scale. Record the main scale reading just before the zero mark of the Vernier scale in Table 1.2. This reading is called main scale reading (M.S.R.).
Note the number (n) of the Vernier scale division which coincides with any division of the main scale.
Repeat the given steps after rotating the Vernier calipers by 90° to measure the internal diameter in a perpendicular direction.
Find total reading and apply zero error correction.

Depth of the beaker

Keep the edge of the main scale of the Vernier caliper on the peripheral edge of the object, as shown in Fig. 1.7.
Keep sliding the moving jaw of the Vernier caliper until the end of the strip just touches the bottom of the beaker.
Repeat the steps given above from four different positions along the circumference of the upper edge of the beaker.
Find total reading and apply zero error correction in Table 1.3.
Take mean of two different values of internal diameter and four different values of the depth.
Calculate the volume by using the formula and show that in the results with proper units.

Observations:

Determination of Vernier constant:

1 small div. of the main scale (MSD) = ______mm or ______cm

_____ Vernier division (VSD) = _____MSD

1 VSD = ______ MSD = ______ mm or ______cm

So, VC = 1 MSD -1 VSD = _______ mm or _______cm

Zero error, e = ± _______mm or _______cm

Table 1.1 Measuring the diameter of spherical body

[(a) and (b) corresponds to mutually perpendicular diameters.]

Mean corrected diameter, D: __________________(cm/mm)

Table 1.2 Determination of internal diameter of a Beaker

[(a) and (b) corresponds to mutually perpendicular diameters.]

Mean corrected diameter of beaker, D: _________________(cm/mm)

Table 1.2 Determination of depth a Beaker

[(A) and (B) corresponds to mutually perpendicular depths.]

Mean corrected depth, h: __________________(cm/mm)

Calculations:

Diameter of the sphere, D = ………………………………cm

Radius of the sphere, r = _____________cm

Internal diameter of beaker, D = …………………………cm

Radius of beaker, r = ________cm

Depth of beaker, h =…………………………………… cm

Volume of beaker, V = ………………………………….cm3

Volume of spherical object, V = ……………………………….cm3

Precautions:

If the motion of the Vernier scale over the main scale is not smooth, lubricate the instrument.
Hold the experimental body firmly but gently between the two jaws of the calipers. Excess pressure may damage the body.
Moreover, if the diameters measured at right angles to each other are found to differ greatly, reject both the values as the body is obviously not uniform along that plane.
Remember, when taking the readings, the line of sight should be normal to the scale where the reading is being taken, in order to avoid parallax error.

Sources of error:

The Vernier scale may be loose.
The jaws may not be at right angle to the main scale.
Parallax error might creep in while taking readings.

Reference:

Feel free to browse through our lab packages, including vernier calipers and physics practical kits.

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Simple Pendulum Experiment Class 11 | Labkafe

Soumen Mondal — Fri, 23 Jul 2021 08:10:33 +0000

Aim:

Using a simple pendulum, plot its L-T2 graph and use it to find the effective length of second’s pendulum.

Apparatus:

A Clamp With Stand
Bob with Hook
Split Cork
Stop Clock/Stop Watch
Vernier Callipers
Cotton Thread
Half Meter Scale

Theory:

A simple pendulum consists of a heavy metallic (brass) sphere with a hook (bob) suspended from a rigid stand, with clamp by a weightless inextensible and perfectly flexible thread through a slit cork, capable of oscillating in a single plane, without any friction, with a small amplitude (less than 150) as shown in figure 6.1 (a). There is no ideal simple pendulum. In practice, we make a simple pendulum by tying a metallic spherical bob to a fine cotton stitching thread.

The spherical bob may be regarded by as a point mass at its centre G. The distance between the point of suspension S and the centre G of the spherical bob is to be regarded as the effective length of the pendulum as shown in figure 6.1 (b). The effective length of a simple pendulum, L = l + h + r. Where l is the length of the thread, h is length of hook, r is radius of bob.

The simple pendulum produces Simple Harmonic Motion (SHM) as the acceleration of the pendulum bob is directly proportional to its displacement from the mean position and is always directed towards it. The time period (T) of a simple pendulum for oscillations of small amplitude, is given by the relation,

T = 2 π √ (L/g)

Where, g = value of acceleration due to gravity and L is the effective length of the pendulum.

T2 = (4π2/g) X L or T2 = KL (K= constant)

and, g = 4π2(L/T2)

If T is plotted along the Y-axis and L along the X-axis, we should get a parabola. If T2 is plotted along the Y- axis and L along the X-axis, we should get a straight line passing through the origin.

Procedure:

Find the vernier constant and zero error of the vernier callipers same as experiment 1.
Measure the radius (r) of the bob using a vernier callipers same as experiment 1.
Measure the length of hook (h) and note it on the table 6.1.
Since h and r is already known, adjust the length of the thread l to make L = l + h + r an integer (say L = 80cm) and mark it as M1 with ink. Making L an integer will make the drawing easier. (You can measure the distance between the point of suspension (ink mark) and the point of contact between the hook and the bob directly. Hence you get l + h directly).
Similarly mark M2, M3, M4 , M5, and M6 on the thread as distance (L) of 90 cm, 100 cm, 110cm, 120cm and 130 cm respectively.
Pass the thread through the two half-pieces of a split cork coming out just from the ink mark (M1).
Tight the split cork between the clamp such that the line of separation of the two pieces of the split cork is at right angles to the line along which the pendulum oscillates.
Fix the clamp in the stand and place it on the table such that the bob is hanging at-least 2 cm above the base of the stand.
Mark a point A on the table (use a chalk) just below the position of bob at rest and draw a straight line BC of 10 cm having a point A at its centre. Over this line bob will oscillate.
Find the least count and the zero error of the stop clock/watch. Bring its hands at zero position
Move the bob by hand to over position B on the right of A and leave. See that the bob returns over line BC. Make sure that bob is not spinning.
Now counting oscillations, from the instant bob passes through its mean position L, where its velocity is maximum. So starting from L it traverses LL2, L2L, LL1, L1L hence, one oscillation is completed. We have to find time for 20 such oscillations.
Now start the stop watch at the instant the bob passes through the mean position A. Go on counting the number of oscillations it completes. As soon as it completes 20 oscillations, stop the watch. Note the time t for 20 oscillations in the table 6.1.
Repeat the measurement at least 3 times for the same length.
Now increase the length of the thread by 10 cm or 15 cm (M2) and measure the time t for this length as explained from step 6 to 14.
Repeat step 15 for at least 4 more different lengths.

Observations:

Vernier constant

Vernier constant of the vernier callipers, V.C. = ______________ cm

Zero error, ±e = _____________cm

Diameter of the bob and length of hook

Observe diameter of the bob:= (i) ______cm, (ii)________cm, (iii)___________cm

Mean diameter of bob, d0 = _________cm

Mean corrected diameter of bob, d = d0 ±e = __________cm

Radius of the bob, r = d/2= ____________ cm

Length of the hook, h= __________cm

Standard value acceleration due to gravity, g1 : 980 cm s-2

Least count of stop clock = ____________s

Zero error of stop clock = ___________s

Table 6.1 Determination of time-periods for different lengths of the pendulum.

Mean = L/T2 = _______________________

Calculation:

We know, T = 2 π √ (L/g)

Experimental value, g1 = 4π2(L/T2) = ______________________

So, %error = (g-g1)/g *100 = ______________________

Graph:

L vs T graph

Plot the graph between L and T from the observations recorded in the table 6.1. Take L along X-axis and T along Y-axis. The L-T curve is a parabola. As shown in the figure 6.2. The origin need not be (0,0) point.

L vs T2 Graph

Plot the graph between L and T2 from the observations recorded in the table 6.1. Take L along X-axis and T2 along Y-axis. The L-T curve is a straight line passing through the (0, 0) point. So the origin of the graph should be chosen (0, 0). As shown in the figure 6.3.

Determination of length of a seconds pendulum from graph:

A second pendulum has time-period 2 s. To find the corresponding length of the pendulum from the L-T graph, draw a line parallel to the L-axis from the point Q1 (0, 2). The line interval the curve L-T at P1. So, the coordinates of P1 is (102, 2).

Length of the seconds pendulum is _____________(102) cm.

To find the length from the L-T2 curve, we, similarly, draw a line parallel to L-axis is form a point Q2 (0, 4). The line intersects the curve at P2. P2 has coordinates (100, 4).

Length of the seconds pendulum is _______________(100) cm.

Precautions :

The thread should be very light and strong.
The point of suspension should be reasonably rigid.
The pendulum should oscillate in the vertical plane without any spin motion.
The floor of the laboratory should not have vibration, which may cause a deviation from the regular oscillation of the pendulum.
The amplitude of vibration should be small (less than 15) .
The length of the pendulum should be as large as possible in the given situation.’
Determination of time for 20 or more oscillations should be carefully taken and repeated for at least three times.
There must not be strong wind blowing during the experiment.

Reference:

http://www.ncert.nic.in/
https://www.learncbse.in/

you may download Simple Pendulum Experiment Class 11 practical manual .pdf Download Now

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Parallelogram Triangle Law of vector addition Experiment Class 11 | Gravesand’s apparatus

Soumen Mondal — Fri, 23 Jul 2021 08:08:26 +0000

Aim:

To find the weight of a given body (Wooden Block) using parallelogram law of vectors.

Apparatus:

Parallelogram Apparatus (Gravesand’s Apparatus)
Two Slotted Weights With Hanger
Wooden Block With Hook
Spring Balance
Mirror Strip
Cotton Thread (roll)
Drawing Pin

Theory :

In Fig. 5.1 we see the Gravesand’s apparatus or Parallelogram apparatus. It consists of a wooden board A fixed vertically on two pillars. There are two pulleys P and Q fitted at the same level at the top of the board. Three set of slotted weights are supplied with the apparatus which can be used to verify the parallelogram law of vectors. A thread carrying hangers for addition of slotted weights is made to pass over the pulleys so that two forces P and Q can be applied by adding weights in the hangers. By suspending the given object, whose weight is to be determined, in the middle of the thread, a third force W is applied.

Fig 5.1

If two vectors acting simultaneously on a particle are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is completely represented in magnitude and direction by the diagonal of that parallelogram drawn from that point.

Let two vectors (P) ⃗ and (Q) ⃗ act simultaneously on a particle O at an angle θ as shown in figure 5.2. They are represented in magnitude and direction by the adjacent sides OA and OB of a parallelogram OACB drawn from a point O. Then the diagonal OC, will represent the resultant R in magnitude and direction. Mathematically we can say,

(R) ⃗= (P) ⃗ + (Q) ⃗

If a body of unknown weight W is suspended from the middle hanger and balancing weights P and Q are suspended from other two hangers then, (R) ⃗ and the three vectors (P) ⃗,(Q) ⃗ ,(W) ⃗ are in equilibrium. Under equilibrium |W| ⃗= |R| ⃗. Weight of a body is a force.

Hence, |W| ⃗ = |P| ⃗+|Q| ⃗.

Equation 1

or,

If S is the actual weight of the body, then the percentage error in the experiment can be calculated using

Equation 2

Procedure :

Set the Gravesand’s apparatus with its board in vertical position with the help of plumb line.
Ensure that the pulleys are moving smoothly. Oil them if necessary.
Fix a drawing paper sheet on the board with the help of drawing pins.
Take a long thread and tie two hangers H1 and H2 to the both ends of the thread.
Tie the given body B, whose weight is to be found, with one end of another shorter thread.
Tie the other end of the shorter thread in the middle of the longer thread to form a small knot, O. This knot becomes the junction of the three threads.
Pass the longer threads over the two pulley P1 and P2 of the apparatus and place two equal slotted weights P and Q the hangers H1 and H2 .
The body B is made to hang vertically with the knot O somewhere in the drawing paper.
Adjust the weights P and Q to keep the knot O at a position slightly below the middle of the paper. This is the equilibrium position of O. Ensure that all the weights hang freely and do not touch the board.
Mark the position of junction O on the white paper by a pencil.
Ensure that there is no friction at the two pulleys P1 and P2. For that purpose, disturb the positions of P and Q a little and leave them. The knot should go back to its initial equilibrium position, If it does not, oil the pulleys to remove friction.
Mark the position of the equilibrium of the knot (O) by a dot of a pencil on the paper sheet.
Take the thin mirror strip m and place it breadthwise on the paper under one of the threads. See the image of the thread in the mirror. Adjust your line of sight such that it is at right angle to strip and the thread. At this point cannot see the image of the thread (meaning, parallax is removed). Mark two points P1 and P2 on the paper, on either side of the strip exactly behind the thread.
Similarly, mark two points Q1, Q2 and W1, W2 for the other two threads.
Remove the threads and hangers from the apparatus and note the weights of the two hanger H1 and H2 along with the slotted weights in table 5.1. Let these two weights be P and Q.
Remove the paper from the board and join P1 and P2, Q1, Q2 are W1, W2 with the help of ruler to get three lines which should meet O. These three lines represent the directions of three forces.
To represent the vector, chose a suitable scale, say 10 gm (0.1N) = 1 cm.
Cut the lines OA and OB to represent the forces and respectively. For example, if P = 60 gm and Q = 60 gm, then OA = 6 cm and OB = 6 cm.
With OA and OB as adjacent sides, complete the parallelogram OACB as shown in fig 5.2.
Join O and C. The length OC represents the resultant vector which corresponds to the unknown weight W.
Measure OC (l) and multiply it by the scale (10 g) to get the value of R.
Measure the angle () using a protractor and calculate W using equation (i).
Repeat the experiment with fresh paper and using different weights (un-equal) P and Q in the hangers.
Find the calculated value of weight R.
Measure the actual weight (S) of the unknown weight (W) by the spring balance and calculate the percentage error using equation (ii).

Observation:

Least count of spring balance, L.C. = _______________ gm

Weight of B by spring balance, S = ________________gm

Scale factor: Let ______ cm = ___________gm

Table 5.1 Measurement of weight of given body

Calculations:

Result:

For equal weights

Weight of unknown body by observation, W: ……………………….

Weight of unknown body by calculation, W‘: ……………………….

For un-equal weights

Weight of unknown body by observation, W: ……………………….

Weight of unknown body by observation, W‘: ……………………….

Precautions :

The pulleys should be frictionless.
The boards should be stable and vertical.
The hangers should not touch the board.
There should be one central knot and it should be small.
When the lines of action of the forces are marked, the hangers should at rest.
The scale should be so chosen that a fairly large parallelogram is obtained.
Threads should be strong and thin, so that they may be regarded as massless.

Reference:

http://www.ncert.nic.in/
https://www.learncbse.in/

You may check out our blog on BEAM BALANCE

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Beam Balance: Physics Experiment Class 11 to determine mass of two different objects | Physical Balance

Soumen Mondal — Fri, 23 Jul 2021 08:04:43 +0000

Aim:

To determine the mass of two different objects using a beam balance.

Apparatus:

Physical Balance
Weight Box
Sprit Level
Two Objects with Different Mass

Theory :

A physical balance or Beam Balance is a weighing instrument that helps in measuring the weight (or gravitational mass) of an object by making use of the principle of moments. It consists of a metal beam B with a knife-edge at the centre pointing in the downward direction. The knife-edge rests on a horizontal flat top made of hard material (Brass). There are two edges with nuts n1 and n2 at the two ends of the metal beam. A pair of pans P1 and P2 are suspended through stirrups S1 and S2 respectively. The nuts n1 and n2 are used to adjust the weights of the pan. In the middle of the beam, there is a long pointer P pointing in the downward direction. This pointer moves on an ivory scale G fixed at the bottom of the brass pillar V. The pillar has two supports K1 and K2 which rests the metal beam when not in use. A knob H, at the bottom of the wooden box, is connected by a horizontal rod to the vertical pillar V. When the handle is rotated rightwards, the beam is raised and is ready to use. There are leveling screws W1 and W2 provided at the bottom of the box to make the pillar horizontal. The plumb line R suspended by the side of the pillar is given to confirm this. There are glass doors provided to the wooden box to avoid the air disturbance and to protect the balance from dust particles present in the air.

When the pans are empty, rotate the handle rightwards. The beam B will rise up and the pointer P will oscillate. If the oscillations are symmetrical about the central division of the ivory scale G, the balanced is under equilibrium state. By adjusting the nuts n1 and n2 ,the instrument can be made in the equilibrium state.

A physical balance determines the gravitational mass of a body by making use of the principle of moments. If the object O having gravitational mass m is placed on one pan and a standard mass O’ of know gravitational mass ms is placed on another pan to keep the beam horizontal, then,

Weight of the body O in one pan = Weight of the body O’ in another pan

or, mg = msg

where g is the acceleration due to gravity, which is constant. Thus,

m = ms

Procedure :

Adjust the physical balance and make it horizontal with the help of spirit level and plumb line. When the beam is in raised position, the pointer C will stay at rest and coincide to the zero division or moves to and fro over the ivory scale G, equally, about the zero division.
When the beam is in the rest position, put any one of the object in the left pan. Now put some standard weight with the help of a forceps from the weight box.
Shut the front glass door so that air current should not disturb. Raise the beam with the help of handle H and notice that beam is horizontal and pointer is moving to and fro equally both sides of zero division.
If, not, then put or remove some fractional weights to get the correct horizontal position of beam and pointer. Bring the beam in rest position and collect all the weight and add them which give the gravitational mass of the object. Now the remove the object also from left pan.
Repeat 2, 3 and 4 for the second object.

Observation:

Table 4.1 Mass of the objects

Sl. No	Object	Weight in gm x	Weight in milligram y	y in gm	Total Weight (x + y) gm
1
2

Result:

The mass of the first object is ……………….. gm.

The mass of the second object is ……………. gm.

Precautions :

Pans must be clean and tidy.
The base should be made horizontal using the levelling screw.
While adding or removing weights, the beam should be at rest.
The body to be weighed should be placed in the left pan weight added to the right pan.
The beam should be raised or lowered gently to avoid damage to the knife-edge.
Masses should always be added in the descending order of magnitude. Masses should be placed in the centre of the pan.
The balance should not be loaded with masses more than capacity. Usually, a physical balance is designed to measure masses upto 250 g.
Weighing of hot and cold bodies using a physical balance should be avoided. Similarly, active substances like chemicals, liquids and powders should not be kept directly on the pan.

Reference:

http://www.ncert.nic.in/
https://www.learncbse.in/

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Screw Gauge experiment class 11 PDF | Micrometer | Least Count | Labkafe

Soumen Mondal — Fri, 23 Jul 2021 07:52:13 +0000

Aim:

To measure (a) diameter of a given wire (b) thickness of a given sheet and (c) volume of an irregular lamina using screw gauge.

Apparatus:

Screw Gauge
Any Wire
Metallic Sheet
Irregular Lamina (uniform thickness)
Millimetre Graph Paper

Theory:

Using Vernier Callipers we can measure length accurately up to 0.1 mm. To measure more accurately, up to 0.01 mm or 0.005 mm, we use screw gauge. A Screw Gauge is an instrument of higher precision than a Vernier Callipers. In any ordinary screw, there are threads and the separation between any two consecutive threads is the same. The distance advanced by the screw when it makes its one complete rotation is the separation between two consecutive threads. This distance is called the Pitch (p) of the screw. It is usually 1 mm or 0.5 mm. Fig. 2.1 shows a screw gauge. It has a screw S which advances forward or backward as one rotates the head C through rachet R. There is a linear scale LS attached to limb D of the U frame.

The smallest division on the linear scale is 1 mm (in one type of screw gauge). There is a circular scale CS on the head, which can be rotated. There are 50 or 100 divisions on the circular scale. When the end B of the screw touches the surface A of the stud/anvil ST, the zero marks on the main scale or pitch scale or linear scale LS and the circular scale should coincide with each other as shown in fig. 2.2.

Principle:

Pitch of the Screw Gauge

The linear distance covered by the tip of the screw (B) in every rotation of the circular scale is called the pitch of a screw gauge. This movement of the spindle is shown on an engraved linear millimeter scale on the sleeve. To find the pitch, give full rotation to the screw (say 4 times) and note the distance (d) advanced by the circular scale over the pitch scale.

If the distance d is 4 mm The pitch can be represented as:

Least Count of the Screw Gauge

On the thimble there is a circular scale which is divided into 50 or 100 equal parts. We are using a screw gauge which has 50 circular divisions. The Least count (LC) is the distance moved by the tip of the screw, when the screw is turned through 1 division of the circular. The least count can be calculated using the formula;

Determination of Zero Error:

When the stud and spindle are brought in contact with each other, the zero of circular scale should coincide with reference line of main scale. In that case the screw gauge have no zero error as shown in Fig. 2.2. However, when the zero of circular scale does not coincide with reference line of main scale, the screw gauge is said to have zero error.

The zero error is said to be positive zero error if on bringing the spindle in contact with stud, if the zero of the circular scale lies to the bottom of the reference line as shown in Fig. 2.3. Owing to this error, the measured readings will be systematically bigger than the actual value by the same amount. Hence the error is to be subtracted from the observed readings. If on the other hand, the zero of the circular scale lies to the top of the reference line as shown in Fig.2.3, it is said to be negative

Fig. 2.2

Fig 2.3

Fig 2.4

zero error. Owing to this error, the measured readings will be systematically smaller than the actual value by the same amount. Hence the error is to be added from the observed readings.

To determine the error, bring the spindle in contact with the stud and note the reading on the linear as well as circular scale. If the linear scale reading is x and circular scale reading is n’ then zero error is given by ± (x + n’ × LC ). Zero correction (e) is always negative of zero error. In our case, as shown in the fig. 2.3, the linear scale reading is zero and the circular scale zerois 2 divisions bellow the reference. Therefore, the zero error is: -[0 + 2 × 0.02] = – 0.04 mm.

So, the Zero correction (e) is = -[-0.04] = 0.04 mm.

Hence, the Actual reading = Measured reading – (±e)

= Measured reading – (-0.04) for positive error

Procedure:

Find the value of one linear scale division (L.S.D.).
Calculate the pitch and the least count of the screw gauge.

Measurement of diameter of the wire

Bring the spindle B in contact with the stud A and calculate the zero error. If there is no zero error, then note down zero error nil.
Move the face B away from face A. Place the wire lengthwise (as shown in the fig.2.5) over face A and move the face B towards face A using the ratchet head R. Stop when R turns (slips) without moving the screw with click sound.
Note the number of divisions of the main scale reading (M.S.R) visible before the edge of circular scale.
Note the number (n) of the division of the circular scale lying over reference line.
Repeat steps 5 and 6 after rotating the wire by 90° for measuring diameter in a perpendicular direction.
Repeat steps 4, 5, 6 and 7 for five different positions separated equally throughout the length of the wire. Record the observations in table 2.1.
Find observed diameter and apply zero correction in each case.
Take mean of different values of actual diameter.

Measurement of thickness of a given sheet

Repeat steps 1, 2, 3, 4, 5 and 6. Instead of wire place the rigid sheer between face A and B.
Find the thickness of the sheet as shown in fig. 2.5 at five different position of the sheet, spread over the surface of the sheet.
Record the observations in the table 2.2.
Find the observed thickness and apply zero correction in each case.
Take mean of different values of actual thickness.

Measurement of volume of an irregular lamina

Repeat steps 1,2,3,4,5 and 6. Instead of wire place the lamina between face A and B.
Find the thickness of the lamina as shown in fig. 2.5 at five different position of the lamina and record them in table 2.3.
Place the lamina on a millimetre graph paper and draw the boundary of the area with a sharp pencil.
Count the number of square enclosed by the boundary. The boundary may contain fractions of many squares. Count those squares, which have fractions greater than half within the boundary as a full squares and ignore those which have less than half within the boundary as shown in the fig 2.6. Naturally there could be some compensation and the result will be very near to the actual value.

Observation:

Determination of least count:

One linear scale division, L.S.D. = ____ mm

Total number of divisions is the in circular scale, N = _______

Distance moved by the screw for 4 rotations, d = ________ mm

Pitch of the screw, p = 4/d = ____mm

Therefore, Least Count, L.C. = p/N= ____mm

Zero error or Instrumental error with the sign:

Zero error, e = ± (x + n’ × LC ) = _______mm

Table 2.1 Determination of diameter of the wire:

No of Obs.	M.S.R (M) mm	No of Circular scale division of the reference line (n)	Observed Diameter D0 = M + n × LC mm	Actual Diameter D = D0 – (± e ) mm
1.(a) (b)
1.(a) (b)
2.(a) (b)
2.(a) (b)
3.(a) (b)
3.(a) (b)
4.(a) (b)
4.(a) (b)
5.(a) (b)
5.(a) (b)

[(a) and (b) corresponds to mutually perpendicular diameters.]

Mean actual diameter, Dw : …………………………………… (mm)

Table 2.2 Determination of thickness of a sheet:

No of Obs.	M.S.R (M) mm	No of Circular scale division of the reference line (n)	Observed Thickness T0 = M + n × LC mm	Actual Thickness T = T0 – (± e ) mm
1.
2.
3.
4.
5.

Mean actual thickness, Ts : …………………………………… (mm)

Table 2.3 Determination of thickness of an irregular lamina:

No of Obs.	M.S.R (M) mm	No of Circular scale division of the reference line (n)	Observed Thickness T0 = M + n × LC mm	Actual Thickness T = T0 – (± e ) mm
1.
2.
3.
4.
5.

Mean actual thickness, TL : …………………………………… (mm)

Calculation:

Number of small squares enclosed by the boundary , NL = _______

Actual Thickness, TL = ___________ mm

Area of the lamina, A = NL × 1 mm2 = _________mm2

Therefore, the volume of the lamina, V = A × TL = …………………………..mm3

Precautions:

The wire should not be pressed tightly between stud and spindle.
Instrumental error should be determined and necessary correction should be taken.
Repeated readings are necessary at different places to ensure uniformity of the wire.
Diameter should be measured in one direction and then in perpendicular direction at the same place, to see whether it is uniform.
Parallax error should take care of.
The Milled head is always to be turned in the same direction, otherwise back-lash error will occur.

Sources of Error:

There might be friction in the screw.
Circular scale divisions may not be equally divided.
There might not be uniformity in the wire.
The sheet and lamina may not be of uniform thickness

Reference:

Your may checkout our blog on HOW TO USE VERNIER CALIPER TO FIND OUT LEAST COUNT AND MEASURE DIAMETER OF SPHERICAL BODY AND BEAKER

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Simple Pendulum Experiment Physics Practical Class 11 and 9 | Labkafe

Soumen Mondal — Fri, 23 Jul 2021 05:03:35 +0000

Pendulum Experiment Physics Practical

Aim:

To study variation of time period of a simple pendulum experiment physics practical of a given length by taking bobs of same size but different masses and interpret the result.

Apparatus:

A Clamp With Stand
Bob with Hook of Different Masses
Split Cork
Stop Clock/Stop Watch
Vernier Callipers
Cotton Thread
Half Meter Scale

Theory:

See Experiment 6.

The time period (T) of a simple pendulum for oscillations of small amplitude, is given by the relation,

T = 2π√(L/g) or T2 = (4π2/g) × L

∴ T ∞√L and T ∞ 1/(√g)

From the above equation, it clearly indicates that the time period of a simple pendulum is independent of mass i.e. for the same value of L and g, the time period of two bobs of different masses will be the same.

Procedure:

Choose any three bobs of known masses and determine their radius as in Experiment 1.
Now, arrange the experiment set up for first bob (say mass m1) with any effective length of simple pendulum (say 90 cm) as explained in Experiment 6. The effective length of the simple pendulum will be kept same in each case.
Record average time taken for 20 or 25 oscillations by the simple pendulum by performing step 4 to 15 as explained in Experiment 6.
Calculate the time periods for each bob and record them in table 7.1.

Observations:

Vernier constant

Vernier constant of the vernier callipers, V.C. = ______________ cm

Zero error, ±e = _____________cm

Least count of stop clock = ____________s

Zero error of stop clock = ___________s

Table 7.1 Determination of time-periods for same lengths of the different pendulum.

Graph:

See Experiment 6.

Result:

For the same value of effective length and acceleration due to gravity, the time period of bobs for different masses are same.

Precautions :

The thread should be very light and strong.
The point of suspension should be reasonably rigid.
The pendulum should oscillate in the vertical plane without any spin motion.
The floor of the laboratory should not have vibration, which may cause a deviation from the regular oscillation of the pendulum.
The amplitude of vibration should be small (less than 15) .
The length of the pendulum should be as large as possible in the given situation.’
Determination of time for 20 or more oscillations should be carefully taken and repeated for at least three times.
There must not be strong wind blowing during the experiment.

Reference:

http://www.ncert.nic.in/
https://www.learncbse.in/

About Labkafe: Lab Equipment Manufacturer & Exporter

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laboratory equipment or lab furniture requirements for school
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