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6.12 Electron mass, me = 9.11*10-31 kg

Proton mass, madhya pradesh = 1.67*10–27 kg

Electron's kinetic energy = 10 keV = 10 $*$ ${10}^3$ $*$ 1.60 *10–19 J = 1.60 *10–15 J

Proton's kinetic energy = 100 keV = 100 $*{10}^3$ $*$ 1.60 *10–19 J = 1.60 *10–14 J

The electron kinetic energy is given by Eke = (1/2)m $v_e^2$ where $v_e$ is the velocity of electron

$v_e$ = $?$  { (2 $*$ Eke )/m} = 5.92 $*{10}^7$ m/s

The velocity of proton $v_p$ = $?$  { (2 $*$ Pke )/m} = 4.37 $*{10}^6$ m/s

The speed ratio = $v_e$ / $v_p$ = 13.5

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6.11: Force exerted on the body, F =?  $\stackrel{?}{i}+2\stackrel{?}{j}$ +3 $\stackrel{?}{k}$ N

Displacement, s = 4 km

Work done, W = F.s

= (?  $\stackrel{?}{i}+2\stackrel{?}{j}$ +3 $\stackrel{?}{k}).(4\stackrel{?}{k})$

= 0+0-3 $*4$

= 12 J

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6.10 Power is given by the relation

P = Fv = mav = mv $\frac{dv}{dt}$ = constant ( say, k)

vdv = $\frac{k}{m}dt$

$\frac{v^2}{2}=\frac{k}{m}t$

v = $\sqrt{\frac{2kt}{m}}$

For displacement x of the body, we have:

v = $\frac{dx}{dt}$ = $\sqrt{\frac{2k}{m}}{t}^{1/2}$

dx = k' $t^{1/2}$ dt where k' = $\sqrt{\frac{2k}{3}}$ = constant

On integrating both sides, we get

x = $\frac{2}{3}k\text{'}{t}^{3/2}$

Therefore x $?t^{3/2}$

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The candidates seeking admission in the G H Raisoni University, Saikheda, BA course are required to visit the official website and fill the application form. The university will release a list of shortlisted students on the basis of Class 12 exam and available number of seats. The document verification is done for the shortlisted students. If there is no discrepancy then the candidates are offered admission.

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6.9 Let us assume

Body mass = m

Acceleration = a

According to Newton's 2nd law F = MA (constant)

We know a = dv/dt = constant. Hence dv = dt $*$ constant

On integrating, v = t + constant

The relation of power is given by P = F $*v$

We have $v=u+at$

Hence,  $P=F*(u+at)$

= $ma(u+at)$

= $mau+m{a}^{2}t$

Therefore P $?t$

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