Let the vertical component of velocity when the bomb strikes the ground be$V_v$

Let the horizontal component of velocity when the bomb strikes the ground be$V_H$

$V_V=O+gt$

Where t = 10 seconds,$V_V=10\times 10=100m/s$and$V_H=500m/s$

$\tan \theta =\frac{V_V}{V_H}=\frac{100}{500}=\frac{1}{5}$

Hence,$\theta ={\tan }^{-1}\left(\frac{1}{5}\right)$

As the density of the liquid is constant, whole of the mass of liquid in the capillary is supposed to be concentrated at the h/2. So, the potential energy of the liquid in capillary is,

$V=\frac{mgh}{2}$

Or,

The mass of the liquid in a column between x and x + dx is
$dm={\mathrm{\pi r}}^2\mathrm{p}\mathrm{dx}$
Therefore, the potential energy of the liquid in a column of height h is,

$\left[\left({\mathrm{\pi r}}^2\right)\times pgdx={\mathrm{\pi r}}^2\mathrm{pg}\right]xdx$

$={\mathrm{\pi r}}^2\mathrm{pg}\frac{{\mathrm{h}}^2}{2}=\left({\mathrm{\pi r}}^2\mathrm{h}\mathrm{p}\right)\mathrm{gh}/2$

= mgh/2
Hence the correct choice is (2).

Here, Z = 8 and A = 17, therefore,

B.E. = (8Mp+ (17 - 8)Mn- Mo)c2

= (8Mp+ 9Mn- Mo)c2

First case
$T^2\infty R^3$
Second Case
$T^2\infty {\left(2R\right)}^3$

$\therefore \frac{T}{T}=\frac{{\left(2R\right)}^{3/2}}{{\left(R\right)}^{3/2}}=2^{3/2}$

$T=2^{3/2}T$
$=2\sqrt{2}\times 24$
$=48\sqrt{2}$hours

Capacitance of parallel plate capacitor with air between the plates is C0= Є0A/d.

When the separation between the plates is reduced to half, C1= Є0A/(d / 2) = 2Є0A/d.

Thus, final capacitance is C2= 10 x 8 pF = 80 pF.

According to Kirchoff's law, current entering the circuit is equal to the current leaving the circuit.

So, 12 + 3 = 18 + I

Or I = - 3A

Let m cells be connected in series and n such groups be connected in parallel. If the ernf of each cell is E and internal resistance r, then the total ernf of m cells in series is mE and the total internal resistance is mr. When n such groups are in parallel, the effective internal resistance is mr/n. Then, the current through an external resistance R is

$I=\frac{mE}{R+{\displaystyle \frac{mr}{n}}}=\frac{mnE}{nR+mr}$

$=\frac{mnE}{{\left(\sqrt{nR}-\sqrt{mr}\right)}^2+2\sqrt{mnRr}}$

Now, I will be maximum if the denominator is minimum, i.e. if nR = mr Given R = 3$\Omega$and r = 1$\Omega$
Using these values, we have 3n = m. But mn = 48
(given). Therefore,$\frac{m\times n}{3}=48,$which m = 12. Thus.

n= 4. Hence, the correct choice is (2).

We know that Q = CV and Q =$Q_o$cos wt. Also,
$Q_o=C{V}_o.$

$\therefore$ $\cos wt=\frac{Q}{Q_o}=\frac{V}{V_o}=\frac{6}{12}=\frac{1}{2}orwt=\frac{\mathrm{\pi }}{3}.$

Now w is given by

$w=\frac{1}{\sqrt{LC}}$ ............(1)

Given$L=0.6\times \times {10}^{-3}HandC=2\times {10}^{-6}F.$Using these values in Eq. (1) we get$w=\frac{{10}^5}{2\sqrt{3}}rad{s}^{-1}.$

Now,$I=\frac{dQ}{dt}=\frac{d}{dt}\left(Q_o\cos wt\right)$

$=-Q_{o}w\sin wt$

$\therefore$ $\left[I\right]=Q_{o}w\sin wt=C{V}_{o}w\sin wt$

$=\left(2\times {10}^{-6}\right)\times 12\times \frac{{10}^5}{2\sqrt{3}}\sin \left(\frac{\mathrm{\pi }}{3}\right)$

$=0.6A$

Since the frequency n of the light does not change as light travels from air into glass, we have,

$v_a=n{\lambda }_{o}and{V}_g=n{\lambda }_g$

Therefore$\frac{{\lambda }_a}{{\lambda }_g}=\frac{V_a}{V_g}=\mu$