JEE 2017 :: Advanced 2017 :: Physics 2017

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

A human body has a surface area of approximately 1m². The normal body temperature is 10K above the surrounding room temperature T₀. Take the room temperature to be T₀= 300K. For T₀  =300K, the value of  $\sigma T_0^4=460 Wm^{-2}$   (where σ is the Stefan Boltzmann constant). Which of the following options 1s/are correct?

Answer:

Explanation:

Assumption e=1

                       [black body radiation]

 $P=\sigma A (T^{4}-T_{0}^{4})$

(c)           $P_{rad}=\sigma AT^{2}=\sigma .1. (T_{0}+10)^{4}$

       =   $\sigma T_0^4(1+\frac{10}{T_{0}})^{4}$

                            [T₀  = 300 Kg given]

    = $\sigma .(300)^{4}(1+\frac{40}{300})=460\times\frac{17}{15}=520J$

     P_net= 520-460=60W

$\Rightarrow$  energy radiated in 1 s=60J

(b)   P=  $\sigma A(T^{4}-T_0^4)$

$dP=\sigma  A (0-4T_0^3 dT)$

  and  $dT= -\triangle T$

 $\Rightarrow$   $dP=4\sigma A T_0^3\triangle T$

(d)   If the surface area decreases, then energy radiation also decreases

Note: While giving answers b and c it is assumed that energy radiated refers to the net radiation. If  energy radiated is taken as the only emission then b and c will not be included in the answer

A block M hangs vertically at the bottom end of a uniform rope constant mass per unit length. The top end of the rope is attached to a fixed rigid support at O. A transverse wave pulse (pulse 1) of wavelength λ₀ is produced at point O on the rope. The pulse takes time T_OA to reach point A. If the wave pulse of wavelength λ ₀  is produced at point A (pulse 2) without disturbing the position of M it takes time T_AO to reach point O. Which of the following options is/are correct?

Answer:

Explanation:

$v=\sqrt{\frac{T}{\mu}}$   so the speed at any position will be same for both pulses, therefore the time taken by both pulses will be same

                    λf=v

                  $\lambda=\frac{v}{f}$

$\Rightarrow$     $λ \propto v \propto T$

since when pulse 1 reaches at A tension and hence speed decreases therefore λ decreases.

A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surfaced of a fixed table. Initially, the right edge of the block is at x=0, in a coordinate system fixed in the table. A point mass m is released from rest at the topmost point of the path as shown and slide down. When the mass loses contact with the block, its position is x and the velocity is v, At that instant, which of the following option is/are correct?

Answer:

Explanation:

$\triangle x_{cm}$ of the block and point mass system=0

    m(x+R)+Mx=0

where x is the displacement of the block

      solving the equation we get

          $x= -\frac{mR}{M+m}$

   From the conservation of momentum and mechanical energy of the combined system.

     0=mv-MV

    $mgR=\frac{1}{2}mv^{2}+\frac{1}{2}MV^{2}$

 Solving these two equations, we get

          $v=\sqrt{\frac{2gR}{1+\frac{m}{M}}}$

A flat plane is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at very low pressure. The speed of the plane v is much less than the average speed u of the gas molecules. Which of the following options is/are true?

Answer:

Explanation:

Just before the collision                            Just after the collision

                   $v_{1}=u+2v$

              $\triangle v_{1}=(2u+2v)$

$F_{1}=\frac{\text{d} \rho _{1}}{\text{d}t}$

   =  $\rho A(u+v)(2u+2v)$

   =  $2 \rho A(u+v)^{2}$

  $v_{2}=(u-2v)$

 $\triangle v_{2}=(2u-2v)$

$F_{2}=\frac{\text{d} \rho _{2}}{\text{d}t}$

= $\rho A(u-v)(2u-2v)$

          $=2 \rho A(u-v)^{2}$

[$\triangle F$   is the net force due to the air molecules on the plate]

      $\triangle F=2 \rho A(4uv)=8 \rho Auv$

    $P= \frac{\triangle F}{A}=8 \rho (uv)$

$F_{net}= (F-\triangle F) =ma$

                         [m is mass of the plate]

   $F -(8 \rho Au)v=ma$

A circular insulated copper wire loop is twisted to form two loops of areas A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field B points into the plane of the paper. At t=0 the loop starts rotating about the common diameter as an axis with a constant angular velocity ω in the magnetic field. Which of the following options is/are correct?

Answer:

Explanation:

The net magnetic flux through that loops at time t is

$\phi = B(2A-A)\cos\omega t=B\omega A \cos\omega t$

So,  $\mid\frac{\text{d}\phi}{\text{d}t}\mid =BA \cos\omega t$

  $\therefore$     $\mid\frac{\text{d}\phi}{\text{d}t}\mid$ is maximum when $\phi=\omega t$ = $\frac{\pi}{2}$

The emf induced in the smaller loop

ε_smaller = $-\frac{\text{d}}{\text{d}t}(BA \cos\omega t)=B\omega A\sin\omega t$

$\therefore$    Amplitude of maximum net induced in  both the loops= Amplitudeof maximum emf induced the smaller loop alone.

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