Class 12th

$A_2{B}_3?2A^{+3}+3B^{-2}$

1-α        2α        3α

$\therefore i=1+4\alpha =1+4*0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4*0.52*1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$         

$logK=logA-\frac{Ea}{2.303RT}$

$Slope=\frac{Ea}{2.303R}=-10,000K$  

$\frac{Ea}{2.303R}=10^4$            

$log\left(10^{-5}\right)=logA-10^4*\frac{1}{500}$   (at 500 K temperature)

$\therefore T=\frac{10^4}{19}K=\frac{100}{19}*10^{2}K=5.2631*10^{2}K=526.31K\approx 526K$

When two or more individual charges are present in a system, the total charge will be an algebraic sum of all individual charges and not the vector sum. Therefore, an electric charge is considered as a scalar quantity.

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

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

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

$X^{\prime }>X$

$M=ml$

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

$=\frac{l}{2}$

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

$=M/2$

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

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

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

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.