VITEEE 2018 :: Physics :: General questions

• Physics - General questions

Four point charges -Q, -q,2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is

Answer:

Explanation:

Let the side length of square be 'a' then potential at centre O is

$V = \frac{k (-Q)}{\left(\frac{a}{\sqrt{2}}\right)}+\frac{k (-q)}{\frac{a}{\sqrt{2}}}+\frac{k (2q)}{\frac{a}{\sqrt{2}}}+\frac{k (2Q)}{\frac{a}{\sqrt{2}}}$ =0

= -Q-q+2q+2Q = 0 → Q+q = 0 → Q = -q

In Young's double slit experiment the two slits are illuminated by light of wavclenght 5890Å and the distance between the fringes obtained on the screen is 0.2º. If the whole apparatus is immersed in water then the angular fringe width will be, if the refractive index of water is 4/3.

Answer:

Explanation:

W_a = λ/d → W_a αλ

$\frac{(W_{a})Water}{W_{a}}$ = $\frac{\lambda_{Water}}{\lambda}$

= $\frac{\lambda}{\mu_{Water}\lambda}$

$(W_{a})Water$ = $\frac{2\times3}{4}$ = 0.15º

A steel wire of length l has a magnetic moment M. It is then bent into a semicircular arc. The new magnetic moment is

Answer:

Explanation:

Let pole strength = m

So, M = ml

When wire is in form of arc, then the distance between poles = $\frac{2l}{\pi}$

So, M' = $\frac{m2l}{\pi}$ = $\frac{2M}{\pi}$

  In the circuit shown the cells A and B have negligible resistances. For V_A 12V,R₁ = 500 Ω and R = 100Ω the galvanometer  (G) shows no deflection. The value of V_B is :

Answer:

Explanation:

Since deflection in galvanometer is zero so current will flow as shown in the above diagram.

Current I = $\frac{V_{A}}{R_{1}+R}$ = $\frac{12}{500+100}$ = $\frac{12}{600}$

So V_B = IR = $\frac{12}{600}\times100$ = 2 V

In a photoelectric effect measurement, the stopping potential for a given metal is found to be V₀ volt when radiation of wavelength λ0 is used.lf radiation of wavelength 2λ₀ is used with the same metal then the stopping potential (in volt) will be

Answer:

Explanation:

$eV_{0} =\frac{hc}{\lambda_{0}}-W_{0}$
and $eV' = \frac{hc}{2\lambda_{0}}-W_{0}$

Subtracting them we have

$e=\left(V_{0}-V'\right) =\frac{hc}{\lambda_{0}}\left[1-\frac{1}{2}\right] =\frac{hc}{2\lambda_{0}}$

or $V' =V_{0}-\frac{hc}{2e\lambda_{0}}$

• Physics - General questions