Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

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 (C_{1}) has 10 equal divisions that correspond to 9 main scale
divisions. The Vernier scale of the other caliper (C_{2}) 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 C_{1} and C_{2} 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

In both calipers C_{1} and C_{2}, 1 cm is divided into 10 equal divisions on the main scale. Thus, 1 division on the main scale is equal to x_{m1} = x_{m2} = 1 cm/10 = 0.1 cm.

In calipers C_{1}, 10 equal divisions on the Vernier scale are equal to 9 main scale divisions.

Thus, 1 division on the Vernier scale of C_{1} is equal to x_{v1} = 9x_{m1}/10 = 0.09 cm.

In calipers C_{2}, 10 equal divisions on the Vernier scale are equal to 11 main scale divisions.

Thus, 1 division on the Vernier scale of C_{2} is equal to x_{v2} = 11x_{m2}/10 = 0.11 cm.

Let main scale reading be MSR and v^{th} division of the Vernier scale coincides with m^{th} division of the main scale (m is counted beyond MSR). The value measured by this calipers is

X = MSR + x = MSR + mx_{m} $$-$$ vx_{v} .........(1)

In calipers C_{1}, MSR_{1} = 2.8 cm, m_{1} = 7 and v_{1} = 7 and in calipers C_{2}, MSR_{2} = 2.8 cm, m_{2} = 8 and v_{2} = 7. Substitute these values in equation (1) to get

X_{1} = MSR_{1} + m_{1}x_{m1} $$-$$ v_{1}x_{v1}

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

X_{2} = MSR_{2} + m_{2}x_{m2} $$-$$ v_{2}x_{v2}

= 2.8 + 8(0.1) $$-$$ 7(0.11) = 2.83 cm.

In calipers C

Thus, 1 division on the Vernier scale of C

In calipers C

Thus, 1 division on the Vernier scale of C

Let main scale reading be MSR and v

X = MSR + x = MSR + mx

In calipers C

X

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

X

= 2.8 + 8(0.1) $$-$$ 7(0.11) = 2.83 cm.

2

MCQ (Single Correct Answer)

Match **List I** with **List II** and select the correct answer using the codes given below the lists:

Q. Coefficient of viscosity

R. Plank Constant

S. Thermal conductivity^{2}T^{-1}]

2. [ML^{-1}T^{-1}]

3. [MLT^{-3}K^{-1}]

4. [ML^{2}T^{-2}K^{-1}]

**List I**

Q. Coefficient of viscosity

R. Plank Constant

S. Thermal conductivity

**List II**

2. [ML

3. [MLT

4. [ML

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

$$\eqalign{
& P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr
& 4\,\,\,\,\,\,1\,\,\,\,\,2\,\,\,\,\,3 \cr} $$

3

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 24^{th} 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

4

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.

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2021 Paper 1 Online (1) *keyboard_arrow_right*

JEE Advanced 2018 Paper 1 Offline (1) *keyboard_arrow_right*

JEE Advanced 2017 Paper 2 Offline (1) *keyboard_arrow_right*

JEE Advanced 2016 Paper 2 Offline (1) *keyboard_arrow_right*

JEE Advanced 2013 Offline (2) *keyboard_arrow_right*

IIT-JEE 2012 (1) *keyboard_arrow_right*

IIT-JEE 2011 (3) *keyboard_arrow_right*

IIT-JEE 2010 (1) *keyboard_arrow_right*

IIT-JEE 2008 (1) *keyboard_arrow_right*

IIT-JEE 2007 (2) *keyboard_arrow_right*

IIT-JEE 2006 (2) *keyboard_arrow_right*

IIT-JEE 2005 Screening (1) *keyboard_arrow_right*

IIT-JEE 2004 Screening (2) *keyboard_arrow_right*

IIT-JEE 2003 Screening (1) *keyboard_arrow_right*

IIT-JEE 2001 Screening (1) *keyboard_arrow_right*

IIT-JEE 2000 Screening (1) *keyboard_arrow_right*

Units & Measurements *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

Work Power & Energy *keyboard_arrow_right*

Impulse & Momentum *keyboard_arrow_right*

Rotational Motion *keyboard_arrow_right*

Gravitation *keyboard_arrow_right*

Properties of Matter *keyboard_arrow_right*

Heat and Thermodynamics *keyboard_arrow_right*

Waves *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

Current Electricity *keyboard_arrow_right*

Electromagnetic Induction *keyboard_arrow_right*

Magnetism *keyboard_arrow_right*

Geometrical Optics *keyboard_arrow_right*

Wave Optics *keyboard_arrow_right*

Modern Physics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*