Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

This is a short answer type question as classified in NCERT Exemplar

Increasing pressure at 0⁰C and 1 atm takes ice into liquid state and decreasing pressure in liquid state at 0⁰C and 1 atm takes to ice state.

When crushed ice is squeezed, some of it melts, filling up the gap between ice flakes upon releasing pressure. This water freezes, binding all ice flakes and making the ball more stable.

This is a short answer type question as classified in NCERT Exemplar

Let the mass and length of a uniform rod be M and l respectively.

Moment of inertia of the rod about its perpendicular bisector I= ml²/12

Increase in length of the rod when temperature is increased by $?$ T is given by

$?l=l\alpha ?T$

So new moment of inertia of the rod = $\frac{M}{12}({l+?l)}^2$ = $\frac{M}{12}\left(l^2+?l^2+2l?l\right)$

As change in length $?l$  is very small therefore neglecting $?l^2$  so we  get

I'= $\frac{M}{12}\left(l^2+2l?l\right)$ = l+mI $\frac{?l}{6}$

Increase in moment of inertia = $\frac{Ml?l}{6}=2*{\left(\frac{m{l}^2}{12}\right)}^1\frac{?l}{l}=2I\alpha ?T$

This is a short answer type question as classified in NCERT Exemplar

As copper is a good conductor of heat as compared to steel. The steel utensils with copper bottom absorbs heat more quickly than steel and give it to food in utensils. So, food heated uniformly and quickly.

This is a short answer type question as classified in NCERT Exemplar

According to the formula  $\frac{F-32}{180}=\frac{C}{100}$

F=C=Q

$\frac{Q-32}{180}=\frac{Q}{100}$ = Q=-40C or 40F

This is a short answer type question as classified in NCERT Exemplar

Due to difference in conductivity metals having high conductivity compared to wood. On touch with a finger heat from the surroundings flows faster to the finger from metals and so one feels the heat.

Similarly when one touches a cold metal the heat from the finger flows away to the surroundings faster.

This is a short answer type question as classified in NCERT Exemplar

$\alpha =\frac{?l}{l?T},$ $\alpha =\frac{4*{10}^{-4}}{2*10}$ = $2*{10}^{-5}$   ^oC^-1

From next observation $?l=2*{10}^{-5}*1*10=2*{10}^{-4}$

From next observation  $?l=2*{10}^{-5}*2*20=8*{10}^{-4}m$

From next observation  $?l=2*{10}^{-5}*3*10$ = $6*{10}^{-4}m$  so this value satisfies the equation.

This is a short answer type question as classified in NCERT Exemplar

As diathermic walls allow exchange of heat energy between two systems and adiabatic walls do not, hence diathermic walls are used to make the bulb of a thermometer.

This is a long answer type question as classified in NCERT Exemplar

(a) Power radiated by stefan's law

P= $\sigma A{T}^4=($ 4 $\pi$ R²)T⁴

= 5.67 $*{10}^{-4}*4*3.14*{0.5}^2*{10}^6*4$

= 1.78 $*{10}^{17}J/s$ =1.8 $*{10}^{17}J/s$

(b) Energy available per second U= 1.8 $*{10}^{17}J/s$ = 18 $*10$ ¹⁶J/s

Actual energy required to evaporate water = 10%of 18 $*{10}^{17}j/s$

=1.8 $*{10}^{16}$ J/s

Energy used per second to raise the temperature of m kg of water from 30⁰C to 100⁰C and then into vapour at 100⁰C

=ms_{w $?\theta$} +mL= m $*4186*\left(100-30\right)+m*22.6*{10}^5$

= 2.93 $*{10}^{5}m+22.6*{10}^{5}m=22.53*{10}^{5}mJ/s$

As per question , 25.53 $*{10}^{5}m$ = 1.8 $*{10}^{16}$

m= $\frac{1.8*{10}^{16}}{25.53*{10}^5}=7*{10}^{9}kg$

(c) Momentum per unit time

P= U/c= $\frac{1.8*{10}^{17}}{3*{10}^8}=6*{10}^{8}kg-m/s$ ²

Momentum per unit time

Area p= p/4 $\pi R^2=\frac{6*{10}^8}{4*3.14*{\left({10}^3\right)}^2}$

d=47.7N/m²

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

Θ= θ₁+ θ₂/2

Let temperature varies linearly in the rod from its one end to other end, let θ be the temperature of the midpoint of the rod. At steady state

Rate of flow of heat,

dQ/dt = $\frac{KA({\theta }_1-\mathrm{\theta })}{\frac{L_o}{2}}=\frac{KA\left(\mathrm{\theta }-{\mathrm{\theta }\mathrm{}}_2\right)}{\frac{L_o}{2}}$

where k is the coefficient of thermal conductivity of the rod

so θ₁- θ= θ- θ₂

θ= θ₁+ θ₂/2

L=L₀ (1+ $\alpha \mathrm{\theta }$ )

L= L_o (1+ $\frac{\alpha \left({\mathrm{\theta }\mathrm{}}_1+{\mathrm{\theta }}_2\right)}{2}$ )

This is a long answer type question as classified in NCERT Exemplar

By applying Pythagoras theorem in given figure

^{${\left(\frac{L+?L}{2}\right)}^2={\left(\frac{L}{2}\right)}^2+x$ 2}

x= $\sqrt[]{{\left(\frac{L+?L}{2}\right)}^2-{\left(\frac{L}{2}\right)}^2}$

=1/2 $\sqrt[]{{(L+?L)}^2-L^2}$

= ½ $\sqrt[]{L^2+{?L}^2+2L?L-L^2}$

=1/2 $\sqrt[]{\left({?L}^2+2L?L\right)}$

As Increase in $?L$  is very small so we neglect it

=1/2 $*\sqrt[]{2?L}$

$?L=L\alpha ?t$

By using this value in above in equation

x=1/2 $*\sqrt[]{2L*L\alpha ?T}$ = ½ L $\sqrt[]{2\alpha ?t}$

= $\frac{10}{2}*\sqrt[]{2*1.2*{10}^{-5}*20}$

= 5 $*\sqrt[]{4*1.2*{10}^{-4}}$

= 5 $*2*1.1*{10}^{-2}=0.11=11cm$