Physics

This is an interesting question! Even though a proton has a positive charge, the net positive charge in a conducting material is always due to the removal of free electrons. 

This happens due to the availability of only free electrons in all conductors. Since protons are present in the nucleus, they can not roam freely in the conductor, as electrons travel freely in the conductor due to their presence in the outer shell.

A magnetic pole will repel or attract magnetic sheet so force is need.

B. If sheet is non-magnetic, no force needed.

C. If it is conducting, then there will be addy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed.

D. The non-conducting and non-polar sheet do not interact with magnetic field of magnet.

$f_0=140 cm$ and $f_e=5 cm$

For distant object,

$m=\frac{f_0}{f_e}=\frac{140}{5}=28$

Given  $V^{\prime }=V=$ Constant

 (i) $C^{\prime }=\frac{{\epsilon }_{0}A}{d^{\prime }},C=\frac{{\epsilon }_{0}A}{d}$

$d^{\prime }<d$

$C^{\prime }>C$
Hence, final capacitance greater than initial capacitance,

(ii)  $U^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2$

$U=\frac{1}{2}C{V}^2$

$U^{\prime }>U$

Hence final energy is greater than initial energy

(iii)  $\frac{Q^{\prime }}{V^{\prime }}=C^{\prime }$ and $\frac{Q}{V}=C$

$\frac{Q^{\prime }}{V^{\prime }}\ne \frac{Q}{V}$

(iv) Product of charge and voltage

$X^{\prime }=Q^{\prime }V=C^{\prime }{V}^2$

$X=QV=C{V}^2$

$X^{\prime }>X$

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially.

After that, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.

Thermal strain $=$ Longitudinal strain $=\alpha \Delta T$

$\Rightarrow$ Longitudinal strain, $\delta =10^{-5}*10^2=10^{-3}$

$\Rightarrow$ Compressive stress $=\delta *$  Young's Modulus

$=10^{-3}*0.5*10^{11}$

$=0.5*10^8$

$\Rightarrow$ Compressive force $=0.5*10^8*10^{-3}=0.5*10^5$

$=5*10^4*\frac{10}{10}$

$=50*10^3 N$

$M=ml$

$\Delta I=2\frac{l}{2}sin30^{?}$

$=\frac{l}{2}$

$M^{\prime }=mI/2$

$=M/2$

Power Consumed $=P=\frac{V^2}{R}$

$\frac{P_A}{P_B}=\frac{R_B}{R_A}$

$R_A=2R_B$

For Series Combination

$P_S=\frac{V^2}{3R_B}$

For Parallel Combination

$P_P=\frac{3V^2}{2R_B}$

$\frac{P_S}{P_P}=\frac{2}{9}$

From principle of homogeneity

$\left[F\right]=\left[\alpha t^2\right]=\left[\beta t\right]$

$\left[\alpha \right]=\frac{\left[F\right]}{\left[t^2\right]} and \left[\beta \right]=\frac{\left[F\right]}{\left[t\right]}$

$\therefore \left[\alpha \right]\left[t\right]=\left[\beta \right]$

$\therefore \frac{\alpha t}{\beta }= dimensionless$

In option (1),

$\frac{10}{15}=\frac{10}{5+R_D}$

The diode can conduct and have resistance  $R_D=10\Omega$  because diode have dynamic resistance. In that case bridge will be balanced.