Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A meter bridge is set up as shown, to determine an unknown resistance X using a standard 10 $$\Omega$$ resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm, respectively, for the ends A and B. The determined value of X is

A

10.2 $$\Omega$$

B

10.6 $$\Omega$$

C

10.8 $$\Omega$$

D

11.1 $$\Omega$$

Corrected length L_{1} (AJ) = 52 + 1 = 53 cm

Corrected length L_{2} (BJ) = (100 $$-$$ 52) + 2 = 50 cm

For a balanced Wheatstone bridge,

$${X \over {10}} = {{{L_1}} \over {{L_2}}} = {{53} \over {50}}$$

$$\Rightarrow$$ X = 10.6 $$\Omega$$

2

MCQ (Single Correct Answer)

A $$2$$ $$\mu F$$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $$S$$ is turned to position $$2$$ is

A

$$0\% $$

B

$$20\% $$

C

$$75\% $$

D

$$80\% $$

When switch S is connected to terminal 1, the potential difference across the 2 $$\mu$$F capacitor is V volt. Therefore, energy stored in the system is

$${U_1} = {1 \over 2}{C_1}{V^2} = {1 \over 2} \times 2 \times {V^2}$$

$$ = {V^2}\,\mu J$$

When switch S is turned to terminal 2, the charge will flow from 2 $$\mu$$F capacitor to 8 $$\mu$$F capacitor until their potentials are equalized. The common potential is

$${V^2} = {q \over {{C_1} + {C_2}}} = {{{C_1}V} \over {{C_1} + {C_2}}}$$

$$ = {{2V} \over {(2 + 8)}} = {V \over 5}$$ volt

$$\therefore$$ Energy stored in the system now will be

$${U_2} = {1 \over 2}({C_1} + {C_2})V_2^2$$

$$ = {1 \over 2}(2 + 8) \times {\left( {{V \over 5}} \right)^2} = {{{V^2}} \over 5}\mu J$$

$$\therefore$$ Percentage loss of energy is

$${{{U_1} - {U_2}} \over {{U_1}}} \times 100 = {{\left( {{V^2} - {{{V^2}} \over 5}} \right)} \over {{V^2}}} \times 100 = 80\% $$

3

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

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

4

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}.

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)

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