VITEEE 2015 :: Physics :: General questions

• Physics - General questions

An electron of mass 9.0 x 10^-31 kg under the action of a magnetic field moves in a  circle of radius 2 cm at a speed of 3 x 10⁶ m/s.  If a  proton  of mass 1.8 x 10²⁷ kg was to move in a circle of same radius in the same magnetic field, then the speed will become

Answer:

Explanation:

Here, the magnetic force (Bqv)  will provide the necessary centripetal force

$\left(\frac{mv^{2}}{r}\right)$

$\therefore$   $Bqv=\frac{mv^{2}}{r}$

 $\Rightarrow$    Bqr= mv

 For electron and proton , the magnetic field  B , charge q and radius  r, all same

 i.e, $m_{e}v_{e}=m_{p}v_{p}$

$v_{p}= (\frac{m_{e}}{m_{p}})v_{e}=\left(\frac{9\times 10^{-31}}{1.8\times 10^{-27}}\right)3\times 10^{6}$.

$=1.5\times 10^{3}m/s$

Two sources of equal emf are connected to a  resistance R. The internal resistance of these sources are r₁ and r₂  (r₁  > r₂). if the potential difference across the source having internal resistance r₂ is zero, then 

Answer:

Explanation:

 Let  emf of each source be E . When they are connected in series , the current in the circuit 

$I= \frac{E_{tot}}{R_{tot}}=\frac{E+E}{r_{1}+r_{2}+R}=\frac{2E}{r_{1}+r_{2}+R}$

 potential  drop across the cell of internal

resistance   $r_{2},\left(\frac{2E}{r_{1}+r_{2}+R}\right)r_{2}$

 Hence,  $E=\frac{2E}{(r_{1}+r_{2}+R)}.r_{2}=0$

$r_{1}+r_{2}+R=2r_{2}$

 $\Rightarrow$  $R=r_{2}-r_{1}$

 In a potentiometer experiment, the balancing length of a cell is 560 cm. When an external resistance of 10 Ω is connected in parallel to the cell, the balancing length changes by 60 cm. The internal resistance of a cell is 

Answer:

Explanation:

 Let us consider a cell of emf E and balancing length l₁

 E=kl₁

 potential difference is balanced by length  l₂ , V= kl₂

 $r= \left(\frac{E-V}{V}\right)R= \left(\frac{E}{V}-1\right)R= \left(\frac{l_{1}}{l_{2}}-1\right)R$

 $r= \left(\frac{560}{560-60}-1\right)10= \left(\frac{56}{50}-1\right)10= \frac{6}{5}=1.2 $ Ω

A 30 V-90W  lamp is operated on a 120 V DC  line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance  is 

Answer:

Explanation:

Resistance of lamb

 $R_{0}= \frac{V^{2}}{P}=\frac{(30)^{2}}{90}=10$ Ω

 Current in lamp

 $I= \frac{V}{R_{0}}=\frac{30}{10}=3 A$

 As the lamp is operated on 120  V DC  , then  resistance becomes

$R'= \frac{V'}{i}=\frac{120}{3}=40$  Ω

 For proper glow , a resistance R is joined in series with the bulb

 R'= R+R₀

 $\Rightarrow  R^{\alpha}=R'- R_{0}=40-10=30$ Ω

The masses of the three copper wires are in the ratio 2:3:5  and their  lengths are in the ratio 5:3:2  . Then , the ratio of their electrical resistance is 

Answer:

Explanation:

Using ,   $R=\rho \frac{l}{A}$

$R_{1 }:R_{2 }:R_{3 }= \frac{l_{1}}{A_{1}}:\frac{l_{2}}{A_{2}}:\frac{l_{3}}{A_{3}}$

$= \frac{l_{1}^{2}}{V_{1}}:\frac{l_{2}^{2}}{V_{2}}:\frac{l_{3}^{2}}{V_{3}}$

$= \frac{l_{1}^{2}}{(m_{1}d)}:\frac{l_{2}^{2}}{(m_{2}d)}:\frac{l_{3}^{2}}{(m_{3}d)}$

$= \frac{l_{1}^{2}}{(m_{1})}:\frac{l_{2}^{2}}{(m_{2})}:\frac{l_{3}^{2}}{(m_{3})}$

$= \frac{5^{2}}{2}:\frac{3^{2}}{3}:\frac{2^{2}}{5}=125:30:8$

• Physics - General questions