TS EAMCET 2017 :: Physics :: General questions

• Physics - General questions

A simple pendulum of length 1m  is freely suspended from the ceiling of an elevator. The time period  of small oscillations as the elevator moves up with an acceleration of 2m/s² is (use g=10 m/s ²)

Answer:

Explanation:

 Given,

 Length of simple pendulum (L) =1m

 Acceleration (a) =2 m/s²

 Gravitation  acceleration (g)=10 m/s²

 We know that,

    $T= 2 \pi \sqrt{\frac{l}{g+a}}=2\pi\sqrt{\frac{1}{10+2}}$

$= 2 \pi \sqrt{\frac{1}{12}}=\pi\sqrt{\frac{4}{12}}$

$I= \frac{\pi}{\sqrt{3}}s$

A coil having n turns and resistance R$\Omega$  is connected with a galvanometer of resistance $4 R\Omega$. This combination is moved in time t seconds from a magnetic flux $\phi_{1}$  Weber to $\phi_{2}$  Weber. The induced current in the circuit is 

Answer:

Explanation:

 Given,

 Number of turns in coil=n

 We know that, 

$Current(i)=\frac{e}{R'}=\frac{-\triangle \phi}{R' \triangle t}$             $\left[\because e=\frac{-\triangle \phi}{\triangle t}\right]$

     $R'= R+4R$

 $i=-\frac{(\phi_{2}-\phi_{1})}{(R+4R)\triangle t}$

$i=\frac{(\phi_{2}-\phi_{1})}{5R(\triangle t)}$

 Here, coil have n number of turns 

Hence, 

$i=-\frac{n(\phi_{2}-\phi_{1})}{5R(\triangle t)}$

 Induced current ,i= $i=-\frac{n(\phi_{2}-\phi_{1})}{5R t}$

A swimmer wants to cross a 200 m wide river which is flowing at a speed of 2 m/s . The velocity of the swimmer with respect to the river is 1 m/s. How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

Answer:

Explanation:

 Given,

 width of river (w)=200 m

 Velocity of river =2 m/s

 Velocity of man =1 m/s

 we know that,

 $\frac{w}{v_{man}}=\frac{d}{v_{stream}}$

 $d= \frac{w \times v_{stream}}{v}$

 $d= \frac{200 \times 2}{1}$

 $d=400 m$

A positive charge Q is placed on a conducting spherical shell with inner radius $R_{1}$, and outer radius $R_{2}$, A particle with charge q is placed at the centre of the spherical cavity. The magnitude of the electric field at a point in the cavity, a distance r from the centre is 

Answer:

Explanation:

 Total charge on shell=+Q

 here, total charge on shell  is +Q . so the (-Q) is on the inner surface of shell.

 Hence, electric field inside the conductor =0

 according the Gauss's law

$\oint Eds=\frac{Q}{\epsilon_{0}}$

$E\oint ds=\frac{Q}{\epsilon_{0}}$

$E(4 \pi r^{2})=\frac{Q}{\epsilon_{0}}$

Electric field,  E=$\frac{Q}{4 \pi  \epsilon_{0} r^{2}}$

 A force F is applied on a square plate of length L. If percentage error in the determination  of L  is 3%  and F is 4%, then the permissible  error  in the calculation of pressure is 

Answer:

Explanation:

 Given,

 Percentage error in length =3%

 Percentage error in force=4%

 We know that,

 Pressure= $P= \frac{F}{A}$

 Area of square = $(L))^{2}$

 $P= \frac{F}{L^{2}}$

$\frac{\triangle P}{p}=\frac{\triangle F}{F}+\frac{2\triangle L}{L}$

$\left(\frac{\triangle P}{p} \times 100\right)=\left(\frac{\triangle F}{F} \times 100\right)+2\left(\frac{\triangle L}{L} \times 100\right)$

 = 4%+2(3%)=10%

• Physics - General questions