VITEEE 2016 :: Physics :: General questions

• Physics - General questions

A current I flows in the anticlockwise direction through a square loop of side a lying in the xoy plane with its centre at the origin. The magnetic induction at the centre of the square loop is 

Answer:

Explanation:

 Field due to one side of loop at O= $\frac{\mu_{0}I}{4\pi (\frac{a}{2})}(2\sin 45^{0})$

 Field at O due to all four sides is along untill vector  $\hat{e}_{x}$

$\therefore$   Total field

 $=4.\frac{\mu_{0}I}{4\pi(\frac{a}{2})}(2\sin 45^{0})= \frac{2\sqrt{2}\mu_{0}I}{\pi a}$

A potentiometer circuit shown in the figure is set up to measure emf of cell E. as the point p moves from X to Y , the galvanometer G shows deflections always in one direction, but the deflection decreases continuously until Y is reached. The balance point between X and Y  may be obtained by

Answer:

Explanation:

Decreasing R increases the current in XY and thereby increases the potential drop across  XP and the balance point may be obtained. The current may be  increased also by increasing V

The current density varies with radial distance r as J = ar², in the cylindrical wire of radius R.The current passing through the wire between radial distance R/3 and R/2 is 

Answer:

Explanation:

Given , J= a r²

$i= \int_{1}^{2}  J\times 2\pi r dr=\int_{R/3}^{R/2}  ar^{2}\times 2\pi r dr$

    =$2\pi a\int_{R/3}^{R/2}  r^{3} dr = 2 \pi a |\frac{r^{4}}{4}|_{R/3}^{R/2}$

$=\frac{\pi a}{2}\left[ (\frac{R}{2})^{4}-(\frac{R}{3})^{4}\right]$

  $=\frac{\pi a R^{4}}{2}\times\frac{65}{81\times 16}$= $\frac{65\pi a R^{4}}{2592}$

Two resistance equal at 0° c with a temperature coefficient of resistance  $\alpha_{1}$  and $\alpha_{2}$ joined in the series act as a single resistance coefficient in a circuit. The temperature coefficient of their single resistance will be

Answer:

Explanation:

$R_{1}=R_{0}(1+\alpha_{1} t)+R_{0}(1+\alpha_{2} t)=2R_{0}\left(1+\frac{\alpha_{1}+\alpha_{2}}{2}t\right)$

 Comparing R= R₀ (1+$\alpha$ t)

$\alpha$ =$\frac{\alpha_{1}+\alpha_{2}}{2}$

 An iron rod of length 2 cm and cross-sectional area of 50 cm² stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. young's modulus of iron rod is 

Answer:

Explanation:

$Y= \frac{F/A}{\triangle l/l}=\frac{\frac{250\times 9.8}{50\times 10^{-6}}}{\frac{0.5\times 10^{-3}}{2}}$

$\frac{250\times 9.8}{50\times 10^{-6}}\times\frac{2}{0.5 \times 10^{-3}}\Rightarrow19.6 \times10^{10} N/m^{2}$

• Physics - General questions