Class 11th

This is a multiple choice answer as classified in NCERT Exemplar

(a) F= V $\rho G$

F $\propto \rho$

$F_{4^{0}C}>F_{0^{0}C}$

Hence buoyancy will be less in water at 0⁰C than that in water at 4⁰C.

This is a multiple choice answer as classified in NCERT Exemplar

(b) Lowest point for scale A is 30⁰ and lowest point for scale B is 0⁰. Highest point for the scale A is 180⁰ and for scale B is 100⁰

$\frac{{\mathrm{t}}_{\mathrm{A}}-{\mathrm{L}\mathrm{E}\mathrm{P}}_{\mathrm{A}}}{{\mathrm{U}\mathrm{F}\mathrm{P}}_{\mathrm{A}}-{\mathrm{L}\mathrm{F}\mathrm{P}}_{\mathrm{A}}}=\frac{{\mathrm{t}}_{\mathrm{B}}-{\mathrm{L}\mathrm{F}\mathrm{P}}_{\mathrm{B}}}{{\mathrm{U}\mathrm{F}\mathrm{P}}_{\mathrm{B}}-{\mathrm{L}\mathrm{F}\mathrm{P}}_{\mathrm{B}}}$

$\frac{{\mathrm{t}}_{\mathrm{A}}-\mathrm{}30}{180-30}=\frac{{\mathrm{t}}_{\mathrm{B}}-0}{100-0}\mathrm{}\mathrm{}$

$\frac{t_A-30}{150}=\frac{t_B}{100}$ ,LFP= lower fixed point , UFP= upper fixed point $\propto \rho$

This is a multiple choice answer as classified in NCERT Exemplar

L = angular momentum = Iw = constant

I₁w₁=I₂w₂

Due to expansion of the rod I₂>I₁

$\frac{w_2}{w_1}=\frac{I_1}{I_2}1$

w₂1 so angular velocity decreases.

This is a multiple choice answer as classified in NCERT Exemplar

(d) As  ${\alpha }_{Al}>{\alpha }_{steel}$ , aluminium will expand more . so it would have larger radius of curvature. So aluminium will expand and will outside region.

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

The first option kept water warmer because according to newton's law of cooling the rate of loss of heat is directly proportional to the difference of temperature of the body and the surroundings and in the first case temperature difference is less. So rate of loss of heat will be less.

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

Mass of water m =100

Change in temperature $?T=0—10=10°C$

Specific heat of water S_w= 1cal/g $°$ C

Latent heat of fusion of water L_fusion= 80cal/g

Heat required to bring water in super cooling from -10 to 0 $°C$

Q= ms $?t=100**10=1000cal$

Let m gram of ice be melted Q= mL

m=Q/L=1000/80= 12.5g

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