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JEE Advanced 2016 Paper 2 Offline

MCQ (Single Correct Answer)
There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2 respectively, are

A
2.85 and 2.82
B
2.87 and 2.83
C
2.87 and 2.86
D
2.87 and 2.87

Explanation

In both calipers C1 and C2, 1 cm is divided into 10 equal divisions on the main scale. Thus, 1 division on the main scale is equal to xm1 = xm2 = 1 cm/10 = 0.1 cm.

In calipers C1, 10 equal divisions on the Vernier scale are equal to 9 main scale divisions.
Thus, 1 division on the Vernier scale of C1 is equal to xv1 = 9xm1/10 = 0.09 cm.

In calipers C2, 10 equal divisions on the Vernier scale are equal to 11 main scale divisions.
Thus, 1 division on the Vernier scale of C2 is equal to xv2 = 11xm2/10 = 0.11 cm.


Let main scale reading be MSR and vth division of the Vernier scale coincides with mth division of the main scale (m is counted beyond MSR). The value measured by this calipers is

X = MSR + x = MSR + mxm $$-$$ vxv .........(1)

In calipers C1, MSR1 = 2.8 cm, m1 = 7 and v1 = 7 and in calipers C2, MSR2 = 2.8 cm, m2 = 8 and v2 = 7. Substitute these values in equation (1) to get

X1 = MSR1 + m1xm1 $$-$$ v1xv1

= 2.8 + 7(0.1) $$-$$ 7(0.09) = 2.87 cm.

X2 = MSR2 + m2xm2 $$-$$ v2xv2

= 2.8 + 8(0.1) $$-$$ 7(0.11) = 2.83 cm.
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JEE Advanced 2013 (Offline)

MCQ (Single Correct Answer)
Match List I with List II and select the correct answer using the codes given below the lists:

List I

P. Boltzmann Constant
Q. Coefficient of viscosity
R. Plank Constant
S. Thermal conductivity

List II

1. [ML2T-1]
2. [ML-1T-1]
3. [MLT-3K-1]
4. [ML2T-2K-1]
A
$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 3\,\,\,\,\,\,1\,\,\,\,\,2\,\,\,\,\,4 \cr} $$
B
$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 3\,\,\,\,\,\,2\,\,\,\,\,1\,\,\,\,\,4 \cr} $$
C
$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 4\,\,\,\,\,\,2\,\,\,\,\,1\,\,\,\,\,3 \cr} $$
D
$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 4\,\,\,\,\,\,1\,\,\,\,\,2\,\,\,\,\,3 \cr} $$
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JEE Advanced 2013 (Offline)

MCQ (Single Correct Answer)
The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
A
5.112 cm
B
5.124 cm
C
5.136 cm
D
5.148 cm
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IIT-JEE 2012

MCQ (Single Correct Answer)
In the determination of Young's modulus $$\left( {Y = {{4MLg} \over {\pi l{d^2}}}} \right)$$ by using Searle's method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement
A
due to the errors in the measurements of d and l are the same.
B
due to the error in the measurement of d is twice that due to the error in the measurement of l.
C
due to the error in the measurement of l is twice that due to the error in the measurement of d.
D
due to the error in the measurement of d is four times that due to the error in the measurement of l.

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