physics ncert solutions class 11th

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

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

Decrease in temperature $?t$ = 57-37= 20⁰C

Coefficient of linear expansion $\alpha$ = 1.7 $*{10}^{-5}/$ ^oC

Bulk modulus for copper B = 140 $*{10}^{9}N/m^2$

Coefficient of cubical expansion $\gamma$ = 3 $\alpha$ = 5.1 $*{10}^{-5}/°C$

Let initial volume of the cavity be V and its volume increases by $?V$  due to increase in temperature.

$?V=\gamma V?t$

$\frac{?V}{V}=\gamma ?t$

Thermal stress produced = B $*volumetricstrain$

= B $*\frac{?V}{V}=B\gamma ?t$

= 140 $*{10}^9*5.1*{10}^{-5}*20$

= 1428 $*{10}^{8}N/m$ ²

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

As difference in volume is constant

By considering the diagram

Let V_io , V_bo be the volume of iron and brass vessel at 0⁰C

V_i,V_b be the volume of iron and brass vessel at $?\theta$ ⁰C

_{${\gamma }_i,{\gamma }_b$} be the coefficient of volume expansion of iron and brass.

V_io -V_bo= 100cc= V_i-V_b

V_i =V_io(1+ $\gamma$ _{i $?\theta$} )

V_b =V_bo(1+ $\gamma$ _{b $?\theta$} )

V_i-V_b = (V_io -V_bo)+ $?\theta \left(V_{io}{\gamma }_i-V_{bo}{\gamma }_b\right)$

Since V_i-V_b= constant

V_{io $\gamma$ i}= V_{bo ${\gamma }_b$}

$\frac{V_{io}}{V_{b0}}=\frac{{\gamma }_b}{{\gamma }_i}=\frac{\frac{3}{2}{\beta }_b}{\frac{3}{2}{\beta }_i}=\frac{{\beta }_b}{{\beta }_i}=\frac{6*{10}^{-5}}{3.55*{10}^{-5}}=\frac{6}{3.55}$

Using above equations

$V_{io}=244.9cc$

V_bo = 144.9cc

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

As l_iron-l_brass =10cm= constant at all temperature

Let l_o be the length of temperature at 0⁰C  and l be the length after change in temperature

l_iron-l_brass =10cm

l_iron (1+ ${\alpha }_{iron}?t$ )-l_brass (1+ ${\alpha }_{brass}?t$ )=10cm

I_{iron $\alpha$ iron}= I_{brass $\alpha$ brass}

$\frac{l_{iron}}{l_{brass}}$ =1.8/1.2=3/2

$\frac{1}{2}{l}_{brass}=10cm$

L_brass=20cm and l_iron=30cm

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Streamline flow is more likely for liquids having low density. We know that greater the coefficient of viscosity of a liquid more will be velocity gradient hence each line of flow can be easily differentiated. Also higher the coefficient of viscosity lower will be Reynolds number, hence flow more like to be streamline.

This is a multiple choice answer as classified in NCERT Exemplar

(c), (d) For liquids coefficient of viscosity $\eta \propto \frac{1}{\sqrt[]{T}}$

i.e with increase in temperature $\eta$ decreases.

For gases coefficient of viscosity $\eta \propto \sqrt[]{T}$

i.e with increase in temperature $\eta$ decreases.

This is a multiple choice answer as classified in NCERT Exemplar

(a), (b) When the coin falls into the water, weight of the (block + coin) system decreases, which was balanced by the upthrust force earlier. As weight of the system decreases, hence upthrust force will also decrease which is only possible when l decreases.

As l decreases volume of water displaced by the block decreases, hence h decreases. As the coin falls into water, it displaces some volume of water which is very less. Hence, we neglect volume of the coin.