JEE 2013 on Main 2013 Questions and Answers | ExamFriend.in | Physics 2013

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The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is

A) 3 V/m

B) 6 V/m

C) 9 V/m

D) 12 V/m

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Answer:

Option B

Explanation:

Peak value of electric field

$E_{0}=B_{0}c= 20\times10^{-9}\times 3\times10^{8}$

=6 V/m

Two coherent point sources S₁ and S₂ are separated by a small distance d as shown. The fringes obtained on the screen will be

A) points

B) staight lines

C) semi-circle

D) concentric circles

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In L-C-R circuit as shown below both switches are open initially. Now switch S₁ and S₂, are closed .(q is charge on the capacitor and

$\tau$ =RC is capacitance time constant) . Which of the following statement is correct?

A) Workdone by the battery is half of the energy dissipated in the resistor

B) At

$t= \tau,q=CV/2$

$t= 2\tau,q=CV(1-e^{-2})$

$t= \frac{\tau}{2},q=CV(1-e^{-1})$

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Answer:

Option C

Explanation:

For charging of capacitor

$q= CV(1-e^{t/\tau})$

at $t=2\tau\Rightarrow q=CV (1-e^{-2})$

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement I: higher the range, the greater is the resistance of ammeter

Statement II: To increase the range of ammeter, additional shunt needs to be used across it.

A) statement I is true , statement II is true and statementII is the correct explanation of statement I

B) Statement I is true , statement II is true, but statement II is not the correct explanation of statement I

C) Statement I is true , statement II is false

D) statement I is false , statement II is true

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10.

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. the piston and the cylinder have equal cross-sectional area A, when the piston is in equilibrium, the volume of the gas is V₀ and its pressure is P₀. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency

$\frac{1}{2\pi}\frac{A_{r}p_{0}}{V_{0b}M}$

$\frac{1}{2\pi}\frac{V_{0}Mp_{0}}{A^{2}\gamma}$

$\frac{1}{2\pi}\sqrt{\frac{A^{2} \gamma p_{0}}{V_{0}M}}$

$\frac{1}{2\pi}\sqrt{\frac{V_{0}M}{A_{\gamma}p_{0}}}$

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Answer:

Option C

Explanation:

In equilibrium

P₀A= Mg when slightly displaced downwards,

$dp=-y(\frac{p_{0}}{V_{0}})dV$

( As in adibatic process, $\frac{dp}{dV}=-y\frac{p}{V}$

Restoring forces,

$F= (dp)A=-\left(\frac{yp_{0}}{V_{0}}\right)(A)(Ax)$

$F\propto -x$

Therefore, motion is simple harmonic comparing with F= -Kx we have:

$K= \frac{yp_{0}A^{2}}{V_{0}}$

$\therefore$ $f=\frac{1}{2\pi} \sqrt{\frac{K}{m}}=\frac{1}{2\pi}\sqrt{\frac{yp_{0}A^{2}}{MV_{0}}}$

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JEE 2013 :: Main 2013 :: Physics 2013

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