Let, mass of bullet = m kg
Rise in temperature of the bullet = \(\triangle\)T⁰C
Velocity of bullet, v = 300 ms^-1
KE. .of bullet = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } m\times 9\times { 10 }^{ 4 }\)
= (4.5 x 10⁴m)J
Useful K.E. to raise the temperature of the bullet
\(w=\frac { 1 }{ 2 } \times 4.5\times { 10 }^{ 4 }m=2.25m\times { 10 }^{ 4 }\)
Heat produced in the bullet, Q = mc \(\triangle\)T
= 0.03 x 4200 m \(\triangle\)T
According to law of conservation of energy
Q =W
\(\therefore\) 126 m \(\triangle\)T = 2.25 m x 10⁴
or \(\triangle\)T = 178.6⁰C

In the first Case
T₁ = 70⁰C, T₂ = 60⁰C, T = 5 min, T₀ = 30⁰C
t = \(\frac { 2.3026 }{ K } \) log₁₀ \(\frac { { T }_{ 1 }-{ T }_{ 0 } }{ { T }_{ 2 }-{ T }_{ 0 } } \)
5 = \(\frac { 2.3026 }{ k } { log }_{ 10 }\frac { 70-30 }{ 60-30 } =\frac { 2.3026 }{ k } \) log₁₀ \(\frac { 4 }{ 3 } \)......(1)
In the Second Case
T₁ = 60⁰C, T₂ = 50⁰C, Tc = 30⁰C
t = \(\frac { 2.3026 }{ k } { log }_{ 10 }\frac { 60-30 }{ 50-30 } \frac { 2.3026 }{ k } { log }_{ 10 }\frac { 3 }{ 2 } \) .......(2)
Dividing (2) by (1), we get
\(\frac { t }{ 5 } =\frac { { log }_{ 10 }1.5 }{ log_{ 10 }1.3333 } =\frac { 0.1761 }{ 0.1249 } \)
t = 1.4 x 5 min = 7 min

For the body,
m₁ = 0.2 kg; \(\triangle\)T₁ = 100⁰ - 16⁰ = 84⁰C
S₁ = ?
For water,
m₂ = 0.5 kg; \(\triangle\)T₂ = 16⁰ - 10⁰ =6⁰C;
S₂ = 4.2 10³J/Kg/⁰C
From law of conservation of energy;
heat lost by body =  heat gained by water
i...e, m₁s₁ \(\triangle\)T₁ = m₂S₂ \(\triangle\)T₂
or S₁ = \(\frac { { m }_{ 2 }s_{ 2 }\triangle { T }_{ 2 } }{ { m }_{ 1 }\triangle { T }_{ 1 } } =\frac { 0.5\times \left( 4.2\times { 10 }^{ 3 } \right) \times 6 }{ 0.2\times 84 } \)
= 0.75 x 10³ J/Kg/⁰C

R_20^0C = 20\(\Omega \), R_40⁰ = 21.2\(\Omega \)
\(\triangle\)\(\theta\) = 40⁰C - 20⁰C = 20⁰C
Using a = \(\frac { { R }_{ 40°C }-{ R }_{ 20°C } }{ { R }_{ 20°C } } \), we get
\(a=\frac { 21.2-20 }{ 20\times 20 } =\frac { 1.2 }{ 400 } \) 
= 3.0 x 10^-30C^-1

If Q be the amount of heat flowing from hot to the cold face, then it is found to be:
(i) directly proportional to the cross-sectional area (A) of the face.
i..e   Q \(\infty\) A
(ii) directly proportional to the temperature difference between the two faces i.e.,
i..e  Q \(\infty\) \(\triangle\)\(\theta\)
(iii) directly proportional to the time t for which the heat flows
i..e., Q \(\infty\) t
(iv) inversely proportional to the distance' d' between the two faces
i..e Q \(\frac { 1 }{ \triangle x } \)
Combining factors (1) to (4), we get
\(Q\infty \frac { A\triangle \theta }{ \triangle x } t\)
or \(Q=KA\frac { \triangle \theta }{ \triangle x } t\)
where K is the proportionality constant known as the coefficient of thermal conductivity

This is because a gas can be heated under different conditions of pressure and volume. The amount of heat required to raise the temperature of unit mass through unit degree is different under different conditions of heating

No. The rate of cooling of a body at a temperature is defined as the fall in temperature per second at that temperature, while the rate of loss of heat is the quantity of heat lost per second from a body at a given temperature.

We know that Q = \(\frac { KA\left( { T }_{ 1 }-{ T }_{ 2 } \right) t }{ l } \)
For given problem , kt = constant or K \(\infty \) \(\frac { 1 }{ t } \)
\(\therefore\)  \(\frac { { k }_{ 1 } }{ { k }_{ 2 } } =\frac { { t }_{ 2 } }{ { t }_{ 1 } } =\frac { 10 }{ 30 } =\frac { 1 }{ 3 } \)

\({ \varepsilon }_{ A }=0.01\ and { \epsilon }_{ B } \ =\ 0.81\)
\(\left( \frac { dQ }{ dt } \right) =\left( \frac { dQ }{ dt } \right) \)
\(\therefore\)  \({ \varepsilon }_{ A }{ \sigma }_{ A }T_{ A }^{ 4 }={ \varepsilon }_{ B }{ \sigma }_{ A }{ T }_{ B }^{ 4 }\)
\(\Rightarrow\) \(\frac { { \varepsilon }_{ A } }{ { \varepsilon }_{ B } } =\frac { { T }_{ B }^{ A } }{ { T }_{ A }^{ 4 } } \)
But \(\lambda \)T = Constant
\(\therefore\)  \(\lambda \)_A\(\lambda \)_A = \(\lambda \)_BT_B
\(\therefore\)  \(\frac { { \varepsilon }_{ A } }{ { \varepsilon }_{ B } } =\frac { { \lambda }_{ A }^{ 4 } }{ { \lambda }_{ B }^{ 4 } } =\frac { 0.1 }{ 0.81 } =\left( \frac { 1 }{ 3 } \right) ^{ 4 }\quad { T }_{ 0 }^{ 4 }\)
\(\frac { { \varepsilon }_{ A } }{ { \varepsilon }_{ B } } =\frac { 1 }{ 3 } \)
\(\lambda \)_B= 3\(\lambda \)_A
\(\therefore\)  \(\lambda \)_B -\(\lambda \)_A = 10^-6
But 3\(\lambda \)_A - \(\lambda \)_A = 10^-6
\(\therefore\) \(\lambda \)_A = 0.5 x 10^-6m
and \(\lambda \)_B = 3 x 0.5 x 10^-6 = 1.5 x 10^-6m

AS radius of ball, r₀ = 9.0 cm = 0.090 m at 0⁰C, hence its
\(Volume,\ \ V_{ 0 }=\frac { 4 }{ 3 } \pi { r }_{ 0 }^{ 3 }=\frac { 4 }{ 3 } \times 3.14\times (0.090)^{ 3 }\)
\(\\ =3.05\times 10^{ -3 }m^{ 3 }\)
\(\\ Again\ as\ \alpha =1.2\times 10^{ -5 }K^{ -1 },\)
\(\\ \therefore \ \gamma =3\alpha \)
\(\\ =3\times 1.2\times 10^{ -5 }=3.6\times 10^{ -5 }K^{ -1 }\)
\(\\ Moreover\ rise\ in\ temperature\)
\(\\ \triangle T=90^{ 0 }C-0^{ 0 }C=90^{ 0 }C=90K\)
\(\\ Increase\ in\ volume,\ \triangle V=V\gamma \triangle T\)
\(\\ =3.05\times 10^{ -3 }\times 3.6\times 10^{ -5 }\times 90\)
\(\\ =9.88\times 10^{ -6 }m^{ 3 }=9.88cm^{ 3 }\) 

The man intends to cut down = 3000 - 2100 = 900kcal
But avoiding sweets completely, he will cut down 700 kcal.To cut down another 200kcal, he should cut down butter by 
\(\frac { 10 }{ 60 } \times 200\simeq 33g\quad per\quad day\)
He should not cut down consumption of food, that provides him with vitamins and other vital nutrients.

(i) As surface area increases on pouring hot tea in sauce from the cup and the rate of loss of heat is directly proportional to surface area of the radiating surface, so the tea will cool faster in the saucer.
(ii) The temperature of hot liquid falls exponentially in accordance with Newton's law of cooling. In other words, rate of the cooling is directly proportional to the temperature difference between hot liquid and the surroundings. It is due to this reason that hot a liquid cools rapidly in the beginning but slowly afterwards.
(iii) When outer surface of container is blackened, the surface  becomes good emitter of heat and so the hot liquid in it cools faster.

0.42 W m^-1K^-1

574.6^o