Thermal Properties of MatterNEET Physics · Class 11 · NCERT Chapter 10

Solution

The heat lost by the steam as it condenses and cools to 80°C is equal to the heat gained by the water. Using the formula $Q=mL+mc\Delta T$, where $m$ is the mass of steam, $L$ is the latent heat of steam, and $c$ is the specific heat of water, we get:
\[ m \times 540 + m \times 1 \times (100 - 80) = 20 \times 1 \times (80 - 10) \]
\[ 540m + 20m = 1400 \]
\[ 560m = 1400 \]
\[ m = \frac{1400}{560} = 2.5\ \text{g} \]
The total mass of water present after the steam condenses is $20\text{g}+2.5\text{g}=22.5\text{g}$, so option (d) is correct.

Solution

Using Newton's law of cooling, $\frac{d\theta }{dt}=k(\theta -{\theta }_0)$. For the first interval, $\frac{70-60}{5}=k(65-{\theta }_0)$, and for the second interval, $\frac{60-54}{5}=k(57-{\theta }_0)$. Solving these, ${\theta }_0=45\text{ºC}$, so option (a) is correct.

Solution

Both alleles are independently expressed in heterozygote during codominance.

Solution

ρ=ρ 0(1−γ∆t) ∆ρ ρ0 =γ∆T= (5× 10−4)(40)=0.02

Solution

λmin T = b λ ∝1 T u ∝(T)4 ∝ 1 (λ)4 So u1 > u2

Solution

Difference in length are same so increase in length are equal ∆𝑙1 = ∆𝑙2 𝑙1𝛼2∆𝑇 = 𝑙2𝛼2∆𝑇 www.vedantu.com 12 ⇒ 𝑙1𝛼1 = 𝑙2𝛼2

Solution

1 4mgh =mL h =4L g = 4 × 3.4 ×105 10 = 13.6 ×104 = 136× 103 km = 136 km

Solution

Thermal current H = H1 + H2 = 112 2 12()()KA T T K A T T dd -- + [] 12 12 12 2( ) ()EQKA T T AT T KKdd - -=+ 12 2EQ KKK +⎡⎤= ⎢⎥⎣⎦

Solution

Rate of power loss 24rR T∝ 24 11 1 24 2 22 rR T r RT = = 14 16× 2 450 1 4r = r2 = 1800 watt

Solution

Using Wien's displacement law, ${\lambda }_0\propto \frac{1}{T}$. If the new wavelength is $\frac{4}{3}{\lambda }_0$, the new temperature $T^{\prime }$ is $\frac{3}{4}T$. The power radiated by a black body is proportional to the fourth power of temperature, $P\propto T^4$. Therefore, the new power $P^{\prime }={\left(\frac{3}{4}\right)}^{4}P=\frac{81}{256}P$. Thus, $n=\frac{256}{81}$, so option (d) is correct.

Solution

The heat added $Q=54\text{cal}=54\times 4.186\text{J}=227.044\text{J}$. The work done by the steam $W=P\Delta V=1.013\times 10^5\times 167.1\times 10^{-6}=17.55\text{J}$. Using the first law of thermodynamics, $\Delta U=Q-W=227.044-17.55=209.494\text{J}\approx 208.7\text{J}$, so option (b) is correct.

Solution

x = 𝐴0 + A sin wt + B cos wt x – 𝐴0 = A sin wt + B cos wt ⸫ Amplitude = √𝐴2+ 𝐵2 + 2𝐴𝐵 𝑐𝑜𝑠90 = √𝐴2+ 𝐵2 2. In which of the following device the eddy current effect is not used? (1) electric heater (2) induction furnace (3) magnetic braking in train (4) electromagnet Solution: (1) Conceptual. 3. Average velocity of a particle executing SHM in one simple vibration is : (1) zero (2) 𝐴ω 2 (3)𝐴ω (4) 𝐴ω2 2 Solution: (1) Average velocity = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 = 0 4. The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t north is given by (1) 45° west (2) 30° west (3) 0° (4) 60° west Solution: (2) 2 1 20 10 mv Rvcos −= − =−= θ θ =120° ∴ angle with north = 30° west 5. In the circuits shown below, the readings of the voltmeters and the ammeters will be: (1)𝑉2> 𝑉1 and 𝑖1 > 𝑖2 (2)𝑉2> 𝑉1 and 𝑖1=𝑖2 (3)𝑉1= 𝑉2 and 𝑖1>𝑖2 (4)𝑉1= 𝑉2 and 𝑖1=𝑖2 Solution: (4) Both voltmeters are ideal. Therefore total current in circuit (2) will flow from upper 10Ω resistance. 6. A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is ( 𝑎𝐶𝑢= 1.7 × 10—5 𝐾—1 and 𝑎𝐴𝑙= 2.2 × 10—5 𝐾—1) (1) 68 cm (2) 6.8 cm (3) 113.9 cm (4) 88 cm Solution: (1) Δl1 = Δl2 l1∝1 Δθ = l2∝2Δθ 88 × 1.7 × 10−5 = l2 × 2.2 × 10−5 Δl2 = 88×1.7/2.2 = 68 cm 7. The unit of thermal conductivity is : (1) W 𝑚—1 𝐾—1 (2) J m 𝐾—1 (3) J 𝑚—1 𝐾—1 (4) Wm 𝐾—1 N E W S 2 vm/s 10 m/s θ 10Ω i1 V1 V1 10V 10Ω i2 A2 V2 10V 10Ω 2 SPACE FOR ROUGH WORK Solution: (1) 𝑄 𝛥𝑡=𝐴𝐾′(𝛥𝜃 𝛥𝑥) (𝐽/𝑠) 𝑚−𝐾=𝐾′ J/s/m/K Or W−m−1−K−1 8. For p-type semiconductor, which of the following statements is true? (1) Electrons are the majority carries and pentavalent atoms are the dopants (2) Electrons are the majority carries and trivalent atoms are the dopants (3) Holes are the majority carri es and trivalent atoms are the dopants (4) Holes are the majority carries and pentavalent atoms are the dopants Solution: (3) Conceptual. 9. A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field. B with the distance d from the centre of the conductor, is correctly represented by figure. (1) (2) (3) (4) Solution: (4) Bin ∝ d (d ≤ R) And 𝐵0∝ 1 𝑑(𝑑>𝑅) 10. Body A of mass 4 m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is: (1) 5 9 (2) 1 9 (3) 8 9 (4) 4 9 Solution: (3) 𝑉1=( 𝑚1−𝑚2 𝑚1+𝑚2 )𝑢1+( 2𝑚2 𝑚1+𝑚2 )𝑢2 𝑣2=( 2𝑚1 𝑚1+𝑚2 )𝑢1+(𝑚2−𝑚1 𝑚1+𝑚2 )𝑢2 Here, v2 = 0 ∴ 𝑣1=( 𝑚1−𝑚2 𝑚1+𝑚2 )𝑣1;𝑣2=( 2𝑚1 𝑚1+𝑚2 )𝑣1 𝑣1=(4𝑚−2𝑚 6𝑚 )𝑢 𝑣1=2𝑚 6𝑚𝑢=1 3 A 4m u B 2m Rest 4m v1 v2 2m Before collision After collision 3 SPACE FOR ROUGH WORK ∴ 𝐾𝑖 = 1 24𝑚𝑢2 and 𝐾𝑓 = 1 24𝑚 𝑢2 9 𝐾𝑖 =2𝑚𝑢2 𝐾𝑓 = 2 9𝑚𝑢2 Loss = 2𝑚𝑢2− 2 9𝑚𝑢2=2𝑚𝑢2× 8 9 ∴ fraction = 2𝑚𝑢2×8 9 2𝑚𝑢2 = 8 9 11. The correct Boolean operation represented by the circuit diagram drawn is: (1) NOR (2) AND (3) OR (4) NAND Solution: (4) Conceptual. 12. When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal. It can travel a distance 𝑥1 along the plane. but when the inclination is decreased to 30° and the same object is shot with the sam e velocity, it can travel 𝑥2 distance. Then 𝑥1∶ 𝑥2 will be : (1) 1 : 2√3 (2) 1 : √2 (3)√2 : 1 (4) 1 : √3 Solution: (4) v2 = u2 + 2a1x1 0 = u2 + 2a1x1 ∴ 𝑥1= −𝑢2 2𝑎1 = −𝑢2 2𝑔𝑠𝑖𝑛θ1 Similarly, 𝑥2= −𝑢2 2𝑔𝑠𝑖𝑛θ2 ∴ 𝑥1 𝑥2 = −𝑢2 2𝑔𝑠𝑖𝑛θ1 × (2𝑔𝑠𝑖𝑛θ2) 𝑢2 = 𝑠𝑖𝑛θ2 𝑠𝑖𝑛θ1 𝑥1 𝑥2 = 1/2 √3/2= 1 √3

Solution

𝐹= 𝑘 𝑄 𝑄 𝑟2 = 𝑘 𝑄2 𝑟2 𝑞1=𝑄−𝑄 4= 3𝑄 4 𝑞1= −𝑄+𝑄 4= −3𝑄 4 ∴ 𝐹1=𝑘 3𝑄 4 × 3𝑄 4𝑟2= 9 16( 𝑘 𝑄2 𝑟2 ) = 9𝐹 16

Solution

The density $\rho$ of an ideal gas is given by $\rho =\frac{P\cdot M}{R\cdot T}$, where $P$ is the pressure, $M$ is the molar mass, $R$ is the gas constant, and $T$ is the temperature in Kelvin. Substituting the values, $P=249\times 10^3\text{ Pa}$, $M=2\times 10^{-3}\text{ kg/mol}$, $R=8.3\text{ J/mol K}$, and $T=27+273=300\text{ K}$, we get $\rho =\frac{249\times 10^3\times 2\times 10^{-3}}{8.3\times 300}\approx 0.2{\text{ kg/m}}^3$. Therefore, option (b) is correct.

Solution

The average thermal energy for a mono-atomic gas is given by $\frac{3}{2}{k}_{B}T$. This is derived from the equipartition theorem, which states that each degree of freedom contributes $\frac{1}{2}{k}_{B}T$ to the energy, and a mono-atomic gas has three translational degrees of freedom. Therefore, option (b) is correct.

Solution

The heat required to raise the temperature of a sphere is proportional to its volume, which is proportional to the cube of its radius. Given $r_1=1.5r_2$, the ratio of volumes is ${\left(\frac{r_1}{r_2}\right)}^3={\left(\frac{1.5r_2}{r_2}\right)}^3=1.5^3=\frac{27}{8}$. Therefore, the ratio of the quantities of heat required is $\frac{27}{8}$, so option (a) is correct.

Solution

Using Newton's law of cooling, $\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$. For the first cooling, $\frac{90-80}{t}=k(90-20)\Rightarrow k=\frac{1}{7t}$. For the second cooling, $\frac{80-60}{T}=k(80-20)\Rightarrow T=\frac{20}{60}\cdot 7t=\frac{10t}{13}$. Thus, the correct option is (c).

Solution

From ideal gas equation PV = nRT mass of water mol. wt.n = 34.5 10 18 ×=   = nRTV P At. STP ⇒ T = 273 K P = 105 N/m2 3 3 5 4.5 10 8.3 273 5.66 m18 10 ××= ×=V - 19 - NEET (UG)-2022 (Code-Q1)

Solution

Path bc is an isochoric process. ∴ Work done by gas along path bc is zero.

Solution

At same temperature, curve with higher volume corresponds to lower pressure. V3 > V2 > V1 ⇒ P1 > P2 > P3 (We draw a straight line parallel to volume axis to get this)

Solution

Sol. In series, $R_{\text{eq.}}=R_1+R_2+R_3$
$=\frac{l}{2KA}+\frac{l}{KA}+\frac{l}{2KA}$
$=\frac{4l}{2KA}$
$R_{\text{eq.}}=\frac{2l}{KA}$

In series rate of heat flow is same
$\therefore \frac{3T-T_1}{R_1}=\frac{3T-T}{R_{\text{eq.}}}$
$\frac{(3T-T_1)2KA}{l}=\frac{(2T)KA}{2l}$
$\Rightarrow 6T-2T_1=T$
$\Rightarrow 2T_1=5T$
$\Rightarrow T_1=\frac{5T}{2}$ ... (1)

Now, equate heat flow rate in 3rd section & total section
$\frac{T_2-T}{R_3}=\frac{3T-T}{R_{\text{eq.}}}$
$\Rightarrow \frac{(T_2-T)(2KA)}{l}=\frac{2T(KA)}{2l}$
$\Rightarrow 2T_2-2T=T$
$\Rightarrow T_2=\frac{3T}{2}$ ... (2)

By equation (1) and equation (2)
$\frac{T_1}{T_2}=\frac{5T\times 2}{2\times 3T}=\frac{5}{3}$