Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the xy-plane and a steady current I flows through the wire. The z-component of the magnetic field at the centre of the spiral is

A

$${{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{b \over a}} \right)$$

B

$${{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{{b + a} \over {b - a}}} \right)$$

C

$${{{\mu _0}NI} \over {2b}}\ln \left( {{b \over a}} \right)$$

D

$${{{\mu _0}NI} \over {2b}}\ln \left( {{{b + a} \over {b - a}}} \right)$$

Magnetic field at the centre of a circular loop of radius r and carrying a current $$I = {{{\mu _0}I} \over {2r}}$$. The direction of the field is along z-direction if the current is anticlockwise.

Consider a small element of width dr. The current through the element is

$$dI = {{total\,current\,in\,spiral} \over {total\,width\,of\,spiral}} \times width\,of\,element$$

$$ = {{Idr} \over {(b - a)}}$$

$$\therefore$$ $$B = \int\limits_a^b {{{{\mu _0}NdI} \over {2r}} = \int\limits_a^b {{{{\mu _0}NI} \over {2(b - a)}}{{dr} \over r}} } $$

$$ = {{{\mu _0}NI} \over {2(b - a)}}\int\limits_a^b {{{dr} \over r} = {{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{b \over a}} \right)} $$

2

MCQ (Single Correct Answer)

Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature T_{c}(0). An interesting property of superconductors is that their critical temperature becomes smaller than T_{c}(0), if they are placed in a magnetic field, that is, the critical temperature T_{c}(B) is a function of the magnetic field strength B. The dependence of T_{c}(B) on B is shown in the figure.

A superconductor has T_{c}(0) = 100 K. When a magnetic field of 7.5 T is applied, its T_{c} decreases to 75 K. For this material, one can definitely say that when

A

B = 5 T, T_{c}(B) = 80 K

B

B = 5 T, 75 K < T_{c} (B) < 100 K

C

B = 10 T, 75 K < T_{c} < 100 K

D

B = 10, T_{c} = 70 K

It is given that T_{c}(0) = 100 K and T_{c}(7.5) = 75 K. Since T_{c}(B) is a monotonically decreasing function of B, T_{c}(5) < T_{c}(0) and T_{c}(5) > T_{c}(7.5). Thus, 75 K < T_{c}(5) < 100 K.

3

MCQ (Single Correct Answer)

Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature T_{c}(0). An interesting property of superconductors is that their critical temperature becomes smaller than T_{c}(0), if they are placed in a magnetic field, that is, the critical temperature T_{c}(B) is a function of the magnetic field strength B. The dependence of T_{c}(B) on B is shown in the figure.

In the graph below, the resistance R of a superconductor is shown as a friction of its temperature T for two different magnetic fields B_{1} (solid line) and B_{2} (dashed line). If B_{2} is larger than B_{1} which of the following graphs shows the correct variation of R with T in these fields?

A

B

C

D

From the given figure, T_{c}(B) is a monotonically decreasing function of B. Thus, B_{2} > B_{1} implies T_{c}(B_{2}) < T_{c}(B_{1}). Hence resistance with B_{2} will become zero at lower temperature in comparison to B_{1}.

4

MCQ (Single Correct Answer)

A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

A

IBL

B

$${{IBL} \over \pi }$$

C

$${{IBL} \over {2\pi }}$$

D

$${{IBL} \over {4\pi }}$$

Consider an small element AB of length dl of the circle of radius R subtending an angle $$\theta$$ at the centre O.

If T is the tension in the wire, then force towards the centre will be equal to $$2T\sin \left( {{\theta \over 2}} \right)$$ which is balanced by outward magnetic force on the current carrying element $$( = IdlB)$$

$$2T\sin \left( {{\theta \over 2}} \right) = IdlB$$

For small angle $$\theta$$, $$\sin {\theta \over 2} \approx {\theta \over 2}$$

or, $$T = {{IBdl} \over \theta } = IBR$$ ($$\because$$ $$\theta = {{dl} \over R}$$)

$$ = {{IBL} \over {2\pi }}$$ ($$\because$$ $$R = {L \over {2\pi }}$$)

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2022 Paper 2 Online (1)

JEE Advanced 2022 Paper 1 Online (1)

JEE Advanced 2021 Paper 2 Online (2)

JEE Advanced 2020 Paper 1 Offline (2)

JEE Advanced 2017 Paper 2 Offline (1)

JEE Advanced 2017 Paper 1 Offline (3)

JEE Advanced 2014 Paper 2 Offline (2)

JEE Advanced 2013 Paper 2 Offline (2)

IIT-JEE 2012 Paper 2 Offline (2)

IIT-JEE 2011 Paper 2 Offline (1)

IIT-JEE 2010 Paper 1 Offline (3)

Units & Measurements

Motion

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