1

JEE Advanced 2015 Paper 1 Offline

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

JEE Advanced 2015 Paper 1 Offline

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

JEE Advanced 2014 Paper 1 Offline

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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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 J1/2 mol$$-$$1/2; $$\sqrt {140RT} $$ = 590 J1/2 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.)
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 $$\left( {M = 36,\sqrt {{{10} \over {36}}} = {{17} \over {32}}} \right)$$

Explanation

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.
4

JEE Advanced 2013 (Offline)

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 90o. The wavelength $$\lambda $$ is exactly known and the error in $$\theta $$ is constant for all values of $$\theta $$. As $$\theta $$ increases from 0o
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

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