Class 11th

11.3 It is given that R = $R_0$ [1 + α(T – $T_0$ )] ………(i)

Where Ro and To are the initial resistance and temperature and R and T are the final resistance and temperature.

At the triple point of water, To = 273.15 K, $R_0$ = 101.6 Ω

At normal melting point of lead, T = 600.5 K, R = 165.5 Ω

Substituting these values in equation (i), we get

165.5 = $101.6$ [1 + α(600.5 – $273.15$ )]

$\frac{165.5}{101.6}$ = 1 + 327.35 α, or α = $\frac{0.629}{327.35}$ = 1.92 $*{10}^{-3}$ $K^{-1}$

When R = 123.4 Ω, T can be calculated as

123.4 = $101.6$ [1 + 1.92 $*{10}^{-3}*$ (T – $273.15$ )]

1.214 = 1 + T1.92 $*{10}^{-3}$ - 1.92 $*{10}^{-3}*$ 273.15

T = 384.6 K

11.2 Triple point of water on absolute scale A,  $T_1$ = 200 A

Triple point of water on absolute scale B,  $T_2$ = 350 B

Triple point of water on absolute Kelvin scale,  $T_k$ = 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 200 on absolute scale A

$T_1=T_k$

200 A = 273.15 K, Therefore A = $\frac{273.15}{200}$

Similarly B = $\frac{273.15}{350}$

If $T_A$ is the triple point of water on scale A and $T_B$ is the triple point of water on scale B, we have

$\frac{273.15}{200}$ $*T_A$ = $\frac{273.15}{350}*$ $T_B$

$T_A=$ $\frac{200}{350}*$ $T_B$ = $\frac{4}{7}*$ $T_B$

11.1 Kelvin and Celsius scales are related as

$T_c$ = $T_k$ - 273.15 ….(i),

Where $T_c=$ temperature in Celsius scale and $T_k$ = temperature in Kelvin scale

Celsius and Fahrenheit scales are related as

$\frac{T_c\mathrm{}}{5}$ = $\frac{T_f-32}{9}$ , where $T_f$ = temperature in Fahrenheit scale

(a) For Neon, $T_k$ = 24.57. Hence $T_c$ = 24.57 – 273.15 = -248.58 degree Celsius

$T_f=\frac{9}{5}*T_c+32$ = $\frac{9}{5}*(-248.58)+32$ = -415.44 degree Fahrenheit

(b) For Carbon dioxide, $T_k$ = 216.55. Hence $T_c$ = 216.55 – 273.15 = -56.6 degree Celsius

$T_f=\frac{9}{5}*T_c+32$ = $\frac{9}{5}*(-56.6)+32$ = -69.88 degree Fahrenheit

Let $\theta$ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction.

mg = N $\cos ?\theta$ …….(1)

mr ${\omega }^2$ = N $\sin ?\theta$ ……(2)

m(R $\sin ?\theta$ ) ${\omega }^2$ = N $\sin ?\theta$

Hence N = m(R) ${\omega }^2$

Substituting the value on N in eqn (1)

mg = mR ${\omega }^2\cos ?\theta$

or $\cos ?\theta$ = g/ R ${\omega }^2$ ………(3)

As $\left|\cos ?\theta \right|$ $\le$ 1, the bead will remain at the lowermost point

g/ R ${\omega }^2$ $\le 1or\omega \le \sqrt{g}/R$

For $\omega$ = $\frac{\sqrt{2g}}{R},equation\left(3\right)$ becomes

$\cos ?\theta$ = g/ R ${\omega }^2$

$\cos ?\theta$ =(g/R)(R/2g) = ½

$\theta =60°$

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction between the wall and his clothing,  $\mu$ = 0.15

Number of revs of hollow cylindrical drum = 200 rev/min = 200/60 rev/s = 3.33 rev/s

The centripetal force required is provided by the normal N of the wall on the man

N = m $v^2/R$ = m ${\omega }^2$ R

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downwards is balanced by the frictional force acting vertically upwards.

The man will not fall, if

mg $\le \mu N$

mg $\le \mu (m{\omega }^{2}R$ )

${\omega }^2\ge g/R\mu$

$g/R\mu$ = 10/ (3 $*$ 0.15)

$\omega =4.71rad/s$

When the motorcyclist is at the uppermost point of the death well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. The outward centrifugal force acting on the motorcyclist is balanced by two forces.

R + mg = m $v^2/r$ , where v is the velocity and m is the combined mass of the motorcycle and motorcyclist

Because of the balance between the forces, the motorcyclist does not fall.

The minimum speed required at the uppermost position to perform a vertical loop is given by R = 0 in the above equation.

So mg = m $v^2/r$ or v = $\sqrt{gr}$ = $\sqrt{10*25}$ = 15.8 m/s

Speed of revolution of the disc, n = $33\frac{1}{3}$ rev/min = 100/3 rpm = 100/ (3 $*60)=0.56rps$

Angular acceleration $\omega$ = 2 $\pi n$ = 2 $*\frac{22}{7}*0.56$ = 3.492 rad/s

The coins revolve with the disc. The centripetal force is provided by the frictional force $m{v}^2/r\le \mu mg$ …. (1)

As v = r $\omega$ , the above equation becomes mr ${\omega }^2/r\le \mu mg$

r $\le \mu g/{\omega }^2$

r $\le$  (0.15 $*10)/{3.492}^2$ = 12 cm

For coin A, r = 4 cm

The condition (r $\le$ 12 ) is satisfied for the coin placed at r = 4 cm, so coin A will revolve with the disc.

The condition (r $\le$ 12 ) is not satisfied for the coin placed at r = 14 cm, so coin B will not revolve with the disc.

Force on the box, F = MA = 40 $*$ 2 N = 80 N

Frictional force, Ff = $?mg$ = 0.15 $*40*10=$ 60 N

Net force = F – Ff = 80 – 60 = 20 N

From the equation F = ma, we get the backward acceleration produced in the box

a = 20/40 = 0.5 m/s2

From the equation s = ut + $\frac{1}{2}a{t}^2$ , to travel s = 5 m by the box to fall off from the truck, we get

5 = 0 $*t$ + $\frac{1}{2}$ $*$ 0.5 $*t^2$

5 = 0.25 $t^2$ , t = 4.47 s

The travel of truck during t = 4.47 s is

= 0 $*t$ + $\frac{1}{2}$ $*$ 0.5 $*t^2$ = 0.5 $*2*{4.47}^2$ = 19.98 m

Mass of the block = 15 kg

Coefficient of static friction between the block and the trolley  $\mu$ = 0.18

Acceleration of the trolley = 0.5 m/s2

(a) Force experienced by block, F = MA = 15 $*$ 0.5 = 7.5 N. This fore acts in the direction of motion of the trolley

Force of friction, Ff = $\mu mg$ = 0.18 $*15*10$ N = 27 N

Force experienced by the block is less than the friction, hence for a stationary observer on the ground, the block will be stationary

(b) When an observer moves with the trolley, the trolley will appear to be at rest