Answer: Option D. -> 1800
:
D
The energy radiated per second by a black body is given by Stefan's Law as $\frac{E}{t}=\sigma T^2\times A$ where A is the surface area of the black body
$\frac{E}{t}\sigma T^4\times 4\pi r^2$
Since black body is a sphere, $(A=4\pi r^2)$.
Case(i)
$\frac{E}{t}=450,T=500K,r=0.12m,$
$\therefore 450=4\pi \sigma \left(500{)}^4\right(0.12{)}^2$ .....(i)
Case(ii)
$\frac{E}{t}=?,T=1000K,r=0.06m,$
dividing (ii) and (I) we get,
$\frac{\frac{E}{t}}{450}=\frac{\left(1000{)}^4\right(0.06{)}^2}{\left(500{)}^4\right(0.12{)}^2}=\frac{2^4}{2^2}=4$
$\Rightarrow \frac{E}{t}=450\times 4=1800W$
(d) is the correct option.

Answer: Option B. -> 6.1 min
:
B
As the temperature differences are small, we can use Newton's law of cooling.
$\frac{d\theta }{dt}=-kt(\theta -{\theta }_0)$
$\frac{d\theta }{\theta -{\theta }_0}=-kdt$
or, ...(i)
Where k is a constant, $\theta$ is the temperature of the body at time t and ${\theta }_0={16}^{\circ }C$ is the temperature of the surrounding. We have,
${\int }_{{40}^{\circ }C}^{{36}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt\left(5min\right)$
$ln\frac{{36}^{\circ }C-{16}^{\circ }C}{{40}^{\circ }C-{16}^{\circ }C}=-kt\left(5min\right)$
or,
$k=-\frac{ln\left(\frac{5}{6}\right)}{5min}$
or, .
If t be the time required for the temperature to fall from 36${}^{\circ }$C to 32${}^{\circ }$C then by (i),
${\int }_{{36}^{\circ }C}^{{32}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt$
or, $ln\frac{{32}^{\circ }C-{16}^{\circ }C}{{36}^{\circ }C-{16}^{\circ }C}=-\frac{ln\left(\frac{5}{6}\right)}{5min}$
or, t=$\frac{ln\left(\frac{4}{5}\right)}{\left(ln\right(\frac{5}{6})}\times 5min$
= 6.1 min.
Alternative method
The mean temperature of the body as it cools from 40${}^{\circ }$C to 36${}^{\circ }$C is $\frac{{40}^{\circ }C+{36}^{\circ }C}{2}={38}^{\circ }C$ The rate of decrease of temperature is
$\frac{{40}^{\circ }C+{36}^{\circ }C}{5min}={0.80}^{\circ }C/min$ .
Newton's law of cooling is
$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$
or, -0.8${}^{\circ }$C/min = - k(38${}^{\circ }$C - 16${}^{\circ }$C)=-k(22${}^{\circ }$C)
or, $k=\frac{0.8}{22}mi{n}^{-1}$.
Let the time taken for the temperature to become 32${}^{\circ }$C be t.
During this period,.
The mean temperature is
Now,

Answer: Option C. -> 955 K
:
C
Let the temperature of the coil be T. The coil will emit radiation at a rate A$\sigma T^4$. Thus,
1000W =$\left(0.020m^2\right)\times (6.0\times {12}^{-8}W/m^2-K^4)\times T64$
or, $T^4=\frac{1000}{0.020\times 6.00\times {10}^{-8}}{K}^4$
=$8.33\times {10}^{11}{K}^4$
or, $T=955K$.

Answer: Option C. -> 0.3
:
C
Before painting it black the power needed = $e\sigma A(T^4-T_0^4)=210$ ...(i)
T = 500 K
$T_0$ = 300 K
e $\to$ emissivity
After the paint the surface behaves like a black body.
$\therefore \sigma A(T^4-T_0^4)=700$ ...(ii)
Dividing (i) and (ii)
$e=\frac{210}{700}$
$\Rightarrow$ e = 0.3

Answer: Option B. -> 887 J
:
B
Stefan's law: $\frac{\Delta Q}{\Delta t}=e\sigma A{T}^4$
For an ideal radiator e = 1 $\Rightarrow \frac{\Delta Q}{\Delta t}=6\times {10}^{-8}\times 1.6\times (273+37{)}^4$
=$6\times {10}^{-8}\times 1.6\times 314\times 104$
= 887 J per second.

Answer: Option A. -> 25681
:
A
According to Wien's law wavelength corresponding to maximum energy decreases. When the temperature of blackbody increases i.e.,
${\lambda }_{m}T=constant\Rightarrow \frac{T_2}{T_1}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{{\lambda }_0}{\frac{3{\lambda }_0}{4}}=\frac{4}{3}$
Now according to Stefan's law
$\frac{E_2}{E_1}={\left(\frac{T_2}{T_1}\right)}^4={\left(\frac{4}{3}\right)}^4=\frac{256}{81}$.

Answer: Option C. -> Spherical
:
C
For all modes of heat transfer $\Delta Q\propto Area$,A.So we should choose the shape with minimum area for minimum heat transfer.
(a)For cubical:
Volume, V = $a^3=1l$,$\therefore a=1dm\left(decameter\right)$
Area, A = $6a^2=6d{m}^2$
(b)For conical: V= $\frac{\pi r^{2}h}{3}$, but given r=h,$r=\frac{3\frac{1}{3}}{\pi }$ dm
A=$\pi r\left(\sqrt{2r}\right)+\pi r^2$
= 7.354 $d{m}^2$
(c)For Sphere:
$V=\frac{4}{3}\pi r^3$, $\therefore r=(\frac{3}{4}{)}^{\frac{1}{3}}dm$
A$=4\pi r^2=4.84d{m}^2$
(d) For Cylinder:
$V=\pi \times r^2\times r\therefore r=\frac{1}{\pi \frac{1}{3}}dm$
A=$2\pi r^2+2\pi r.r=5.86d{m}^2$
As Sphere has the least area among these, it will be the shape with minimum heat loss. In fact sphere is the shape with the minimum surface area among all possible shapes!

Answer: Option C. -> 19.1 hours
:
C
$t=\frac{Ql}{KA({\theta }_1-{\theta }_2)}=\frac{mLl}{KA({\theta }_1-{\theta }_2)}=\frac{V\rho Ll}{KA({\theta }_1-{\theta }_2)}$
$=\frac{5\times A\times 0.92\times 80\times \frac{5+10}{2}}{0.004\times A\times 10\times 3600}=19.1hours.$

Answer: Option B. -> A and B will represent cooling curves of water and oil respectively
:
B

As we know, Rate of cooling $\propto \frac{1}{specificheat\left(c\right)}$
$\because C_{oil}<C_{water}$
$\Rightarrow (Rateofcooling{)}_{oil}>(Rateofcooling{)}_{water}$
It is clear that, at a particular time after start cooling, temperature of oil will be less than that of water.
So graph B represents the cooling curve of oil and A represents the cooling curve of water.

Answer: Option C. -> 6.9×10−3kg
:
C
Heat transferred in one minute is utilised in melting the ice so, $\frac{KA({\theta }_1-{\theta }_2)t}{l}=m\times L$
$\Rightarrow m=\frac{{10}^{-3}\times 92\times (100-0)\times 60}{1\times 8\times {10}^4}=6.9\times {10}^{-3}kg$