1

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
$$L \propto \sqrt G $$
2

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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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.
3

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
4

IIT-JEE 2010

MCQ (More than One Correct Answer)
A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?
A
Error ΔT in measuring T, the time period, is 0.05 seconds
B
Error ΔT in measuring T, the time period, is 1 second
C
Percentage error in the determination of g is 5%
D
Percentage error in the determination of g is 2.5%

Joint Entrance Examination

JEE Advanced JEE Main

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

Medical

NEET

CBSE

Class 12