physics ncert solutions class 11th

6.21 Given, the area of the windmill sweep = A, Wind velocity = v

The volume of air passing through the blade = Av

Let the density of air be $?$ , the mass of air passing through the blade = $?Av$

(a) The mass of air passing through the blade in time t = $?Avt$

(b) The kinetic energy of air = $\left(\frac{1}{2}\right)m{v}^2$ = $\left(\frac{1}{2}\right)?Avt{v}^2$ = $\left(?At{v}^3\right)$ /2 …. (1)

(c) Area, A = 30 $m^2$ , v = 36 km/h = 10 m/s, density of air be $?$ = 1.2 kg/ $m^3$

Total wind energy, from eqn. (1) = 18 kW

Electrical energy = 25 % of wind energy = 0.25 $*18=4.5kW$

6.20 Mass of the body = 0.5 kg

Velocity, v = a x 3/2

a = 5 m–1/2 s–1

At x =0, the initial velocity, u = 0

At x = 2, the final velocity, v = 5 $*2^{3/2}$ = 14.142 m/s

Work done by the system = increase in K.E. of the body = (1/2)m ( $v^2$ - $u^2$ )

= (1/2) $*0.5*$ 14.142 $*14.142$ = 50 J

6.19 Given, the mass of the trolley,  $m_t$ = 300 kg, mass of the sand bag,  $m_s$ = 25 kg, uniform velocity of the trolley, v = 27 kmph = 0.75 m/s

Since there is no external force acting on the system, the speed of the trolley will remain unchanged even after entire sand is empty. 27 kmph is the answer.

6.18 The length of the pendulum, l = 1.5 m

The potential energy of the bob at horizontal position = mgl

Since it dissipates 5% of its kinetic energy to come to the horizontal position, from the law of conservation of energy we get,

$\left(\frac{1}{2}\right)m{v}^2$ = (0.95) $*mgl$

$v^2$ = 2 $*0.95*9.81*1.5$

v = 5.287 m/s

6.17 In an elastic collision, when the ball A will hit the ball B, A comes to rest immediately and the ball B acquires the velocity of ball A. The momentum thus gets transferred from a moving body to a stationary body.

6.16 The mass of the ball bearing = m

Before the collision, the total K.E. of the system = K.E. of the stationary ball bearing + K.E. of the striking ball bearing = 0 + $\left(\frac{1}{2}\right){mv}^2$

After the collision, the K.E. of the total system is

(a) Case (i) = 0 + $\left(\frac{1}{2}\right)\left(2m\right){\left(\frac{v}{2}\right)}^2$ = $\left(\frac{1}{4}\right){mv}^2$

(b) Case (ii) = 0 + $\left(\frac{1}{2}\right)m{v}^2$

(c) Case (iii) = $\left(\frac{1}{2}\right)\left(3m\right){\left(\frac{v}{3}\right)/}^2$ = $\left(\frac{1}{6}\right)m{v}^2$

Case (ii) is possible since K.E. is conserved in this case.

6.15 Given, the volume of the tank, V = 30 $m^3$

Time required to fill the tank, t = 15 min = 900 s

Height of the tank above the ground, h = 40 m

The efficiency of the pump? = 30%

The density of water,  $?$ = ${10}^3$ kg/ $m^3$

Now, the mass of the water pumped, m = $?V$ = 30 $*{10}^3$ kg = 3 $*{10}^4$ kg

Power consumed = W/t = mgh/t = 3 $*{10}^4*9.8*$ 40 / 900 = 13066 W

P input = Power consumed /? = 43.6 kW

6.14 Momentum is always conserved for an elastic or inelastic collision.

The molecule's initial velocity, u = final velocity v = 200 m/s

Initial kinetic energy = (1/2)m $u^2$

Final kinetic energy = (1/2)m $v^2$ = (1/2)m $u^2$

Therefore, kinetic energy is also conserved

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