Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Consider a thin square sheet of side L and thickness, made of a material of resistivity $$\rho$$. The resistance between two opposite faces, shown by the shaded areas in the figure is

A

directly proportional to L.

B

directly proportional to t.

C

independent to L.

D

independent of t.

The sheet is of square shape with thickness t, width $$\omega$$ = L, and length l = L. The resistance between the two opposite faces is given by

$$R = {{\rho l} \over A} = {{\rho l} \over {wt}} = {{\rho L} \over {Lt}} = {\rho \over t}$$.

2

MCQ (Single Correct Answer)

To verify Ohm's law, a student is provided with a test resitor R_{T}, a high resistance R_{1}, a small resistance R_{2}, two identical galvanometers G_{1} and G_{2}, and a variable voltage source V. The correct circuit to carry out the experiment is

A

B

C

D

To verify Ohm's law, we need to measure the voltage across the test resistance R_{T} and current passing through it. The voltage can be measured by connecting a high resistance R_{1} in series with galvanometer. This combination becomes a voltmeter and should be connected in parallel to R_{T}. The current can be measured by connecting a low resistance R_{2} (shunt) in parallel with galvanometer. This combination becomes an ammeter and should be connected in series to measure the current through R_{T}.

3

MCQ (Single Correct Answer)

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100, 60 and 40 W bulbs have filament resistances R_{100}, R_{60} and R_{40} respectively, the relation between these resistances is

A

$${1 \over {{R_{100}}}} = {1 \over {{R_{40}}}} + {1 \over {{R_{60}}}}$$

B

$${R_{100}} = {R_{40}} + {R_{60}}$$

C

$${R_{100}} > {R_{60}} > {R_{40}}$$

D

$${1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$

The power of the bulb is

$$P = {{{V^2}} \over R}$$

Therefore,

$$100 = {{{V^2}} \over {{R_{100}}}} \Rightarrow {1 \over {{R_{100}}}} = {{100} \over {{V^2}}}$$

where R_{100} is the resistance (at any temperature) corresponds to 100 W. Similarly,

$$60 = {{{V^2}} \over {{R_{60}}}} \Rightarrow {1 \over {{R_{60}}}} = {{60} \over {{V^2}}}$$ and $$40 = {{{V^2}} \over {{R_{40}}}} \Rightarrow {1 \over {{R_{40}}}} = {{40} \over {{V^2}}}$$

From these equations, we get

$${P_{100}} > {P_{60}} > {P_{40}} \Rightarrow {1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2016 Paper 2 Offline (2)

JEE Advanced 2016 Paper 1 Offline (1)

JEE Advanced 2014 Paper 2 Offline (1)

JEE Advanced 2013 Paper 2 Offline (2)

IIT-JEE 2012 Paper 2 Offline (1)

IIT-JEE 2012 Paper 1 Offline (1)

IIT-JEE 2011 Paper 1 Offline (2)

IIT-JEE 2010 Paper 1 Offline (3)

Units & Measurements

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Laws of Motion

Work Power & Energy

Simple Harmonic Motion

Impulse & Momentum

Rotational Motion

Gravitation

Properties of Matter

Heat and Thermodynamics

Waves

Wave Optics

Geometrical Optics

Electrostatics

Current Electricity

Magnetism

Electromagnetic Induction

Alternating Current

Dual Nature of Radiation

Atoms and Nuclei