JEE 2017 :: Advanced 2017 :: Physics 2017

• Advanced 2017 - Physics 2017
• Advanced 2017 - Chemistry 2017
• Advanced 2017 - Mathematics 2017

The instantaneous voltage at three terminals marked X, Y and Z are given by V_X= V₀ sinωt.

$V_{Y}=V_{0}\sin(\omega t+\frac{2\pi}{3}) and$

$V_{Z}=V_{0}\sin(\omega t+\frac{4\pi}{3})$

 An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z . The reading (s) of the voltmeter will be

Answer:

Explanation:

$V_{XY}=V_{0}\sin(\omega t+\frac{2\pi}{3})-V_{0}\sin\omega t$

  = $V_{0}\sin(\omega t+\frac{2\pi}{3})+V_{0}\sin(\omega t+\pi)$

$\Rightarrow$   $\phi =\pi -\frac{2\pi}{3}=\frac{\pi}{3}$

$\Rightarrow$   $V_{0}^{'}= 2V_{0}\cos(\frac{\pi}{6})$

            = $\sqrt{3}V_{0}$

$\Rightarrow$    $V_{XY}=\sqrt{3}V_{0}\sin (\omega t+\phi)$

$(V_{XY})_{rms}= (V_{YZ})_{rms}=\sqrt{3}\frac{V_{0}}{\sqrt{2}}$

A uniform magnetic field B exists in the region between x=0 and $x=\frac{3R}{2}$

 (region 2 in the figure)pointing normally into the plane of the paper. A particle with charge +Q and momentum p directed along X-axis enters region 2 from region 1 at point P₁(y=-R)

which of the following option(s) is/are correct?

Answer:

Explanation:

a)

(b)

  r(1-cosθ)=R

$r\sin\theta =\frac{3R}{2}$

$\frac{\sin\theta}{1-\cos\theta}=\frac{3}{2}$

$\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}}=\frac{3}{2}$

$\cot\frac{\theta}{2}=\frac{3}{2}$

$\tan\frac{\theta}{2}=\frac{2}{3}$

$\tan\theta=\frac{2(\frac{2}{3})}{1-\frac{4}{9}}=\frac{\frac{4}{3}}{\frac{5}{9}}=\frac{12}{5}$

$\sin\theta=\frac{12}{13}$

$r(\frac{12}{13})=\frac{3R}{2}; r=\frac{13R}{8}=\frac{P}{QB};B=\frac{8P}{13QR}$

(c) $\frac{P}{QB}<\frac{3R}{2},B>\frac{2P}{3QR}$

(d)  $r=\frac{mv}{QB},d=2r=\frac{2mv}{QB}\Rightarrow d\propto m$

A source of constant voltage V is connected to a resistance R and two ideal inductors L₁ and L₂ through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t=0, the switch is closed and current begins to flow. Which of the following options is/are correct?

Answer:

Explanation:

since inductors are connected in parallel

$V_{L_{1}}=V_{L_{2}}$

$L_{1}\frac{dI_{1}}{dt}=L_{2}\frac{dI_{2}}{dt}$

$L_{1}I_{1}=L_{2}I_{2}$

$\frac{I_{1}}{I_{2}}=\frac{L_{2}}{L_{1}}$

Current through resistor  at any time t is given by

$I= \frac{V}{R}(1-e^{-\frac{RT}{L}}) where L=\frac{L_{1}L_{2}}{L_{1}+L_{2}}$

 after log time   $I= \frac{V}{R}$

                    I₁ +I₂ =I            .........(i)

                    L₁I₁  = L₂I₂       ........(ii)

From eq(i) and (ii) , we get

  $I_{1}=\frac{V}{R}\frac{L_{2}}{L_{1}+L_{2}}$,  $I_{2}=\frac{V}{R}\frac{L_{1}}{L_{1}+L_{2}}$

(d) value of current is zero at t=0

value of current V/R at t=∞

 hence option (d) is incorrect

A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque  $\tau$ about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct ?

Answer:

Explanation:

(a) If force is applied normal to surface at  P, then the line of action of force will pass from  Q and thus $\tau$ =0

(b) wheel can climb

(c) $\tau=F(2R\cos\theta)-mgR\cos\theta$

             $\tau\propto\cos\theta$

Hence, as θ increases, $\tau$ decreases so it correct

(d)

$  \tau =Fr_{\perp}-mg\cos\theta:\tau$ increases with θ

A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is θ. Which of the following statements about its motion is/are correct?

Answer:

Explanation:

when the bar makes an angle of the height of its COM (midpoint)  $\frac{L}{2}\cos \theta$

$\therefore$    Displacement

                           =$L-\frac{L}{2}\sin\theta=\frac{L}{2}(1-\cos\theta)$

Since, force on COM is only along the vertical direction, Hence COM is falling vertically downwards instantaneous  torque about point of contact is

  $\tau =mg \times \frac{L}{2}\sin\theta$   or $\tau\propto sin\theta$

Now,

                     $x=\frac{L}{2}\sin\theta$

               $y=L\cos\theta$

$\frac{X^{2}}{(\frac{L}{2})^{2}}+\frac{Y^{2}}{L^{2}}=1$

  Path of A is an ellipse

• Advanced 2017 - Physics 2017
• Advanced 2017 - Chemistry 2017
• Advanced 2017 - Mathematics 2017