Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (More than One Correct Answer)

Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a
screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier
scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the
circular scale moves it by two divisions on the linear scale. Then:

A

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the
screw gauge is 0.01 mm.

B

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the
screw gauge is 0.005 mm.

C

If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,
the least count of the screw gauge is 0.01 mm.

D

If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,
the least count of the screw gauge is 0.005 mm.

2

MCQ (More than One Correct Answer)

Planck's constant h, speed of light c and gravitational constant G are used to form a unit of length L and a
unit of mass M. Then the correct option(s) is(are)

A

$$M \propto \sqrt c $$

B

$$M \propto \sqrt G $$

C

$$L \propto \sqrt h $$

D

$$L \propto \sqrt G $$

3

MCQ (More than One Correct Answer)

A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s^{$$-$$1}. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 $$\pm$$ 0.005) m, the gas in the tube is

(Useful information : $$\sqrt {167RT} $$ = 640 J^{1/2} mol^{$$-$$1/2}; $$\sqrt {140RT} $$ = 590 J^{1/2} mol^{$$-$$1/2}. The molar mass M in grams is given in the options. Take the values of $$\sqrt {10/M} $$ for each gas as given there.)

(Useful information : $$\sqrt {167RT} $$ = 640 J

A

Neon $$\left( {M = 20,\sqrt {{{10} \over {20}}} = {7 \over {10}}} \right)$$

B

Nitrogen $$\left( {M = 28,\sqrt {{{10} \over {28}}} = {3 \over 5}} \right)$$

C

Oxygen $$\left( {M = 32,\sqrt {{{10} \over {32}}} = {9 \over {16}}} \right)$$

D

Argon $$\left( {M = 36,\sqrt {{{10} \over {36}}} = {{17} \over {32}}} \right)$$

Minimum length $$ = {\lambda \over 4} \Rightarrow \lambda = 4l$$

Now, v = f $$\lambda$$ = (244) $$\times$$ 4 $$\times$$ l

as l = 0.350 $$\pm$$ 0.005

$$\Rightarrow$$ v lies between 336.7 m/s to 346.5 m/s

Now, $$v = \sqrt {{{\gamma RT} \over {M \times {{10}^{ - 3}}}}} $$, here M is molecular mass in gram $$ = \sqrt {100\gamma RT} \times \sqrt {{{10} \over m}} $$.

For monoatomic gas,

$$\gamma$$ = 1.67

$$\Rightarrow$$ $$v = 640 \times \sqrt {{{10} \over m}} $$

For diatomic gas,

$$\gamma = 1.4 \Rightarrow v = 590 \times \sqrt {{{10} \over m}} $$

$$\therefore$$ $${v_{Ne}} = 640 \times {7 \over {10}} = 448$$ m/s

$${v_{Ar}} = 640 \times {{17} \over {32}} = 340$$ m/s

$${v_{{O_2}}} = 590 \times {9 \over {16}}=331.8$$ m/s

$${v_{{N_2}}} = 590 \times {3 \over 5} = 354$$ m/s

$$\therefore$$ Only possible answer is Argon.

Now, v = f $$\lambda$$ = (244) $$\times$$ 4 $$\times$$ l

as l = 0.350 $$\pm$$ 0.005

$$\Rightarrow$$ v lies between 336.7 m/s to 346.5 m/s

Now, $$v = \sqrt {{{\gamma RT} \over {M \times {{10}^{ - 3}}}}} $$, here M is molecular mass in gram $$ = \sqrt {100\gamma RT} \times \sqrt {{{10} \over m}} $$.

For monoatomic gas,

$$\gamma$$ = 1.67

$$\Rightarrow$$ $$v = 640 \times \sqrt {{{10} \over m}} $$

For diatomic gas,

$$\gamma = 1.4 \Rightarrow v = 590 \times \sqrt {{{10} \over m}} $$

$$\therefore$$ $${v_{Ne}} = 640 \times {7 \over {10}} = 448$$ m/s

$${v_{Ar}} = 640 \times {{17} \over {32}} = 340$$ m/s

$${v_{{O_2}}} = 590 \times {9 \over {16}}=331.8$$ m/s

$${v_{{N_2}}} = 590 \times {3 \over 5} = 354$$ m/s

$$\therefore$$ Only possible answer is Argon.

4

MCQ (More than One Correct Answer)

Using the expression $$2d\sin \theta = \lambda $$, one calculates the values of d by measuring the corresponding angles $$\theta $$ in
the range 0 to 90^{o}. The wavelength $$\lambda $$ is exactly known and the error in $$\theta $$ is constant for all values of $$\theta $$. As $$\theta $$ increases from 0^{o}

A

the absolute error in d remains constant

B

the absolute error in d increases

C

the fractional error in d remains constant

D

the fractional error in d decreases

On those following papers in MCQ (Multiple Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2021 Paper 2 Online (1) *keyboard_arrow_right*

JEE Advanced 2020 Paper 1 Offline (1) *keyboard_arrow_right*

JEE Advanced 2019 Paper 1 Offline (1) *keyboard_arrow_right*

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

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

JEE Advanced 2015 Paper 2 Offline (1) *keyboard_arrow_right*

JEE Advanced 2015 Paper 1 Offline (2) *keyboard_arrow_right*

JEE Advanced 2014 Paper 1 Offline (1) *keyboard_arrow_right*

JEE Advanced 2013 Offline (1) *keyboard_arrow_right*

IIT-JEE 2010 (1) *keyboard_arrow_right*

IIT-JEE 1998 (2) *keyboard_arrow_right*

IIT-JEE 1995 Screening (1) *keyboard_arrow_right*

IIT-JEE 1986 (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*