JEE 2016 :: Advanced 2016 :: Physics 2016

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016

A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x₀. Consider two cases : (i) when the block is at x₀  and (ii) when the block is at x=x₀ +A . In both the cases, a particle with mass m (<M)  is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M?

Answer:

Explanation:

Case-1

Case -II

In caes-I

$Mv_{1}=(M+m)v_{2}\Rightarrow v_{2}=(\frac{M}{M+m})v_{1}$

$\sqrt{\frac{k}{M+m}}A_{2}=(\frac{M}{M+m})\sqrt{\frac{k}{M}}A_{1}$

$A_{2}=\sqrt{\frac{k}{M+m}}A_{1}$

In case-2,     $A_{2}=A_{1}$

$T=2\pi \sqrt{\frac{M+m}{k}}$ in both cases.

 Total energy decreases in first case whereas remain same in 2nd case. Instantaneous speed at x₀ decreases in both cases.

In an experiment to determine the acceleration due to  gravity g, the formula used for the  time  period of a periodic motion is 

$T=2\pi\sqrt{\frac{7(R-r)}{5g}}$ . The values of R and r are measured to be (60±1) mm and   (10± 1) mm respectively. In five successive measurements. the time period is found to be 0.52s, 0.56s, 0.57s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01s, Which of the following statement(s) is (are)  true?

Answer:

Explanation:

Mean time period

$=\frac{0.52+0.56+0.57+0.54+0.59}{5}$

 =0.556=0.56 sec as per significant figures

 Error in reading=| T_mean- T₁|=0.04

$| T_{mean}-T_{2}|=0.00\Rightarrow|T_{mean}-T_{3}|=0.01$

$| T_{mean}-T_{4}|=0.02\Rightarrow|T_{mean}-T_{5}|=0.03$

 Mean error=0.1/5=0.02

% of errot in T= $\frac{\triangle T}{T}\times 100=\frac{0.02}{0.56}\times100 =3.57$ %

% error in r= $\frac{0.001 \times 100}{0.010}=10$ %

% error in R=  $\frac{0.010 \times 100}{0.600}=1.67$

% error in $\frac{\triangle g}{g}\times 100$

=$\frac{\triangle (R-r)}{R-r}\times 100+2\times\frac{\triangle T}{T}$

$=\frac{0.002\times 100}{0.05}+2\times 3.57$

= 4%+7%=11%

Consider two identical galvanometer and two identical resistors with resistance R. If the internal resistance  of the galvanometers

R_c < R/2, which of the following statement(s) about anyone is (are) true? 

Answer:

Explanation:

The maximum range of voltage resistance should be maximum , So all four sholud be connected in series, For maximum range of current , net resistance should be least. Therefore, all four should be connected in parallel

Two thin circular discs of mass m and 4m having radii of a and 2a , respectively are rigidly fixed by a massless, rigid rod of length 

l=$\sqrt{24}a$ through their centers, This assemply is laid on a firm and flat surface and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω .The angular momentum of the entire assemply about the point 'O' is L (see the figure). Which of the following statement(s) is (are) true?

Answer:

Explanation:

$\cos\theta =\frac{1}{\sqrt{l^{2}+a^{2}}}=\frac{\sqrt{24}}{5}$

(c) Velcocity of point P:aω=1Ω then

Ω = $\frac{a\omega}{1}=$ Angular velocity of C M w.r.t

Point O. Angular velocity fo CM w.r.t Z - axis = Ω cosθ

$\omega_{CM-Z}=\frac{a\omega}{1}\frac{\sqrt{24}}{5}=\frac{a\omega}{\sqrt{24}a}\frac{\sqrt{24}}{5}$

$\omega_{CM-Z}=\frac{a\omega}{5}$

(a) $L_{z}=L_{CM}-0^{\cos\theta}-L_{D}-_{CM}sin\theta$

$=\frac{81\sqrt{24}}{5}a^{2}m\omega \times\frac{\sqrt{24}}{5}-\frac{17ma^{2}\omega}{2}\times\frac{1}{\sqrt{24}}$

$=\frac{81\times 24 ma^{2}\omega}{25}-\frac{17ma^{2}\omega}{2\sqrt{24}}$

(b) $L_{CM-O}=(5m)[\frac{9l}{5}Ω]\frac{9l}{5}=\frac{81ml^{2}\omega}{5}$

  $=\frac{81ml^{2}}{5}\times\frac{a\omega}{l}$

$L_{CM-O}=\frac{81 mla\omega}{5}=\frac{81\sqrt{24}a^{2}m\omega}{5}$

(d)  $L_{D-CM}=\frac{ma^{2}}{2}\omega+\frac{4m(2a)^{2}}{2}\omega=\frac{17ma^{2}\omega}{2}$

A small object is placed 50cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle θ=30°  to the axis of the lens, as shown in figure

 If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x,y) at which the image is formed are

Answer:

Explanation:

For Lens

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow v=\frac{uf}{u+f}$

$v=\frac{(-50)(30)}{-50+30}=75cm$

For Mirror

   $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow v=\frac{u f}{u-f}$

$\Rightarrow v=\frac{(\frac{25\sqrt{3}}{2})(50)}{\frac{25\sqrt{3}}{2}-50}=\frac{-50\sqrt{3}}{4-\sqrt{3}}cm$

$m=-\frac{v}{u}=\frac{h_{2}}{h_{1}}$

$\Rightarrow h_{2}=-[\frac{\frac{-50\sqrt{3}}{4-\sqrt{3}}}{\frac{25\sqrt{3}}{2}}]\frac{25}{2}\Rightarrow h_{2}=\frac{+50}{4-\sqrt{3}}$

The x-coordinate of the images

$=50-v \cos 30+h_{2} \cos 60 \approx 25$

The v-coordinate of the images

$=v\sin 30+h_{2}\sin60 \approx 25\sqrt{3}$

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016