VITEEE 2017 :: Physics :: General questions

• Physics - General questions

A barmagnet of magneticmoment M, is placed in a magnetic field of induction B. The torque exerted on it is 

Answer:

Explanation:

$\tau=MB \sin\theta=\vec{M}\times\vec{B}$

Two capacitors when connected in series have a capacitance of 3 μF and when connected in parallel have a capacitance of 16 μF. Their individual capacities are

Answer:

Explanation:

$C_{s}= \frac{C_{1}C_{2}}{C_{1}+C_{2}} =3$

$C_{p}=C_{1}+C_{2} = 16 \therefore C_{1}C_{2}=48$

$C_{1}-C_{2} = \sqrt{(C_{1}+C_{2})^{2}-4C_{1}C_{2}}$ $= \sqrt{16^{2}-4\times48}$=8

C₁+C₂ = 16μF

C₁-C₂=8μF

2C₁ =  24μF = C₁ = 12μF

.'. C₂ = 48/12 = 4μF

In young's double-slit experiment, the intensity of tight at a point on the screen where the path difference is λ, is I, λ being the wavelength of light used. The intensity at a point where the path difference is λ/4 will be

Answer:

Explanation:

For path difference λ, phase difference = $ 2\pi$  ($Q = \left(\frac{2\pi}{\lambda}x=\frac{2\pi}{\lambda}\lambda=2\pi\right)$)

$I = I_{0}+I_{0}+2I_{0}\cos2\pi$

$I = 4I_{0} (\therefore\cos 2\pi = 1)$

For $x= \frac{\lambda}{4}$, Phase difference = $\frac{\pi}{2}$

$I' = I_{1}+I_{2}+2\sqrt{I_{1}}\sqrt{I_{2}}\cos\pi/2.$

$If = I_{1} = I_{2} = I_{0}$ then I' = $2I_{0}$ = 2.I/4 = I/2

A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be

Answer:

Explanation:

A galvanometer is converted into an ammeter, by connected a shunt, in parallel with it.

$\frac{GS}{G+S}=\frac{V_{G}}{I}=\frac{25\times10^{-3}}{25}$

$\frac{GS}{G+S}=0.001$Ω

Here S<<G so S = 0.001 Ω

Resistances 1Ω, 2Ω and 3Ω are connected to form a triangle. If a 1.5 V cell of negligible internal resistance is connected across the 3 Ω resistor, the current flowing through this resistor will be

Answer:

Explanation:

Equivalent resistance between A and B = series combination of 1Ω and 2Ω in paralell with 3Ω resistor

$\frac{1}{R}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ or R = 1.5 Ω

.'. Current in the circuit is I = V/R = 1.5/1.5 = 1A

Since the resistance in arm ACB = resistance in arm AB = 3 Ω, the current divides equally in the two arms. Hence the current through the 3 Ω resistor = I/2 = 0.5 A.

• Physics - General questions