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In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is 

Answer:

Explanation:

Induced constant . $I= \frac{e}{R}$

 Here, e= induced emf

 $= \frac{d\phi}{dt}$

 $I=\frac{e}{R}= (\frac{d\phi}{dt}).\frac{1}{R}$

 $d\phi=IRdt$

 $\phi =\int_{}^{} IRdt$

 $\therefore$   Here, R is constant

  $\therefore$     $\phi =R\int_{}^{} Idt$

$\int_{}^{} I.dt =$  Area under I- t graph

 = $\frac{1}{2}\times 10 \times 0.5 =2.5$

 $\therefore$     $\phi =R\times 25$

                          = 100 × 2.5

                          =250 Wb

The temperature of an open room of volume 30 m³  increases from 17° C to 27° C due to the sunshine . The atmosphere pressure in the room remains 1× 10⁵ Pa. If n_i and n_f  are the number of molecules in the room before and after heating , then n_f  -n_i will be

Answer:

Explanation:

From $pV=nRT=\frac{N}{N_{A}}RT$

    We have, $n_{f}-n_{i}=\frac{pVN_{A}}{RT_{f}}-\frac{pVN_{A}}{RT_{i}}$

$\Rightarrow n_{f}-n_{i}=\frac{10^{5}\times30}{8.30}\times6.02\times 10^{23}$

    $(\frac{1}{300}-\frac{1}{290})$

   $=-2.5\times 10^{25}$

$\therefore$     $\triangle n =-2.5\times 10^{25}$

A diverging lens with a magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of the magnitude of focal length 20 cm. A  beam of parallel light falls on the diverging lens. The final image formed is

Answer:

Explanation:

Focal length of the diverging lens is 25 cm.

   As the rays are coming parallel, so the image (I₁) will be formed at the focus of diverging lens  i.e., at 25 cm towards left of diverging lens

Now, the image (I₁) will work as an object for a converging lens.

       For a converging lens, the distance of object u (i.e, the distance of I₁)  = -(25+15)

                                                                                                              = -40 cm

                                                                      f  = 20 cm

$\therefore$       From lens 's formula  

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

  $\frac{1}{20}= \frac{1}{v}-\frac{1}{-40}\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{40}$

    $\frac{1}{v}= \frac{1}{90}\Rightarrow v =40 cm$

 v is positive so image will be real and will form at right side of converging lens at 40 cm.

An external pressure P is applied on a cube at 0° C . so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

Answer:

Explanation:

$K = \frac{P}{(-\triangle V/V)}$

  $\Rightarrow   -\frac{\triangle V}{V}=\frac{P}{K}$

$\Rightarrow   -\triangle V=\frac{PV}{K}$

change in volume $\triangle V= \gamma V \triangle T$ 

  where $\gamma$ = coefficient of volume expansion

  Again , $\gamma = 3\alpha$

  $\alpha$ is coefficient of linear expansion

$\therefore$       Δ V =V (3 $\alpha$)ΔT

$\therefore    \frac{PV}{K}=V (3\alpha)\triangle T$

$\therefore    \triangle T= \frac{P}{3\alpha K}$

A  slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see the figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is 

Answer:

Explanation:

As the rod rotates in the vertical plane so torque is acting on it, which is due to the vertical component of the weight of the rod.

Now, Torque $\tau$ =force × perpendicular distance of the line of action of force from axis of rotation

                                           =  $mg\sin\theta \times\frac{1}{2}$

    Again , Torque $\tau= Iα$

Where, I = moment of interia =  $\frac{ml^{2}}{3}$

      [ Force and Torque frequency along axis of rotation passing through in end ]

$\alpha$ =  angular acceleration

  $\therefore$    $mg\sin\theta \times \frac{1}{2}=\frac{ml^{2}}{3}\alpha$

  $\therefore$    $\alpha   =\frac{3g\sin\theta}{2l}$

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