JEE Temp-Physics Questions

The mean free time $$\tau$$ between two successive collisions is obtained from the relation

$$\tau = \frac{\lambda}{\bar{c}}$$

where $$\lambda$$ is the mean free path and $$\bar{c}$$ is the average (root-mean-square) speed of the molecules.

The collision frequency $$f$$, i.e. the number of collisions per second, is the reciprocal of the mean free time:

$$f = \frac{1}{\tau} = \frac{\bar{c}}{\lambda}$$

Substituting the given values $$\lambda = 3 \times 10^{-7}\,\text{m}$$ and $$\bar{c} = 600\,\text{m s}^{-1}$$:

$$f = \frac{600}{3 \times 10^{-7}}\;\text{s}^{-1}$$

$$f = \frac{600}{3}\times 10^{7}\;\text{s}^{-1}$$

$$f = 200 \times 10^{7}\;\text{s}^{-1}$$

$$f = 2 \times 10^{9}\;\text{s}^{-1}$$

Therefore, the collision frequency of oxygen molecules at 300 K and 1 atm is $$2 \times 10^{9}\,\text{s}^{-1}$$, which matches Option C.

For gases at same temperature,

$$v_{rms}=\sqrt{\frac{3RT}{M}}$$

So

$$v_{rms}\propto\frac{1}{\sqrt{M}}$$

Therefore

Molar masses:

$$M_{H_2}=2$$

$$M_{O_2}=32$$

Given

$$v_{H_2}=2\ \text{km s}^{-1}$$

So

2vO2=322\frac{2}{v_{O_2}}

=

\sqrt{\frac{32}{2}}vO2​​2​=232​​

$$=\sqrt{16}=4$$

Thus

$$v_{O_2}=\frac{2}{4}$$

$$=0.5\ \text{km s}^{-1}$$

The collision frequency of a gas molecule (number of collisions per unit time for a single molecule) is given by

$$f = \sqrt{2}\;\pi d^2 n \bar{v}$$

where $$d$$ is the molecular diameter, $$n$$ is the number density, and $$\bar{v}$$ is the mean speed.

Since the temperature remains constant, both $$\bar{v}$$ and $$d$$ remain the same (because mean speed depends only on temperature and molecular mass). So $$f \propto n$$.

The number density increases from $$n_1 = 3 \times 10^{19}\;\text{cm}^{-3}$$ to $$n_2 = 12 \times 10^{19}\;\text{cm}^{-3}$$. The ratio of collision frequency before and after is then

$$\frac{f_1}{f_2} = \frac{n_1}{n_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = \frac{1}{4} = 0.25$$

Hence, the correct answer is 0.25.

The rms speed and average speed of gas molecules are given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}, \quad v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$

The ratio:

$$\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3RT/M}{8RT/(\pi M)}} = \sqrt{\frac{3\pi}{8}}$$

Given: $$v_{rms} = \left(1 + \frac{5}{x}\right)^{1/2} \cdot v_{avg}$$

Squaring both sides:

$$\frac{v_{rms}^2}{v_{avg}^2} = 1 + \frac{5}{x} = \frac{3\pi}{8}$$

$$\frac{5}{x} = \frac{3\pi}{8} - 1 = \frac{3\pi - 8}{8}$$

$$x = \frac{40}{3\pi - 8}$$

Using $$\pi = \frac{22}{7}$$:

$$3\pi - 8 = \frac{66}{7} - \frac{56}{7} = \frac{10}{7}$$

$$x = \frac{40}{\frac{10}{7}} = \frac{40 \times 7}{10} = 28$$

Therefore, the correct answer is Option C: $$\mathbf{28}$$.

We need to convert $$-95°$$X to the Fahrenheit scale, using Celsius as an intermediate.

On the X scale, the freezing point is $$-15°$$X and the boiling point is $$65°$$X, giving a range of $$65 - (-15) = 80$$ divisions. The conversion formula from X to Celsius is:

$$\frac{T_X - (-15)}{80} = \frac{T_C - 0}{100}$$

Substituting $$T_X = -95$$:

$$\frac{-95 + 15}{80} = \frac{T_C}{100}$$

$$\frac{-80}{80} = \frac{T_C}{100}$$

$$T_C = -100°\text{C}$$

Now, converting to Fahrenheit:

$$T_F = \frac{9}{5}T_C + 32 = \frac{9}{5}(-100) + 32 = -180 + 32 = -148°\text{F}$$

So, the answer is $$-148°$$F.

Use the general relation for any linear temperature scales:

$$\frac{\text{Reading}-\text{Lower fixed point}}{\text{Upper fixed point}-\text{Lower fixed point}}=\text{constant}$$

For scale P:

• Lower fixed point = 30
• Upper fixed point = 180

For scale Q:

• Lower fixed point = 0
• Upper fixed point = 100

So,

$$\frac{t_P-30}{180-30}=\frac{t_Q-0}{100-0}$$

$$\frac{t_P-30}{150}=\frac{t_Q}{100}$$

The r.m.s. speed of a gas molecule is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

At the same temperature, the ratio of r.m.s. speeds of two gases is:

$$\frac{v_{Ar}}{v_{Cl_2}} = \sqrt{\frac{M_{Cl_2}}{M_{Ar}}}$$

Given: $$v_{Cl_2} = 490$$ m/s, $$M_{Cl_2} = 70.9$$ u, $$M_{Ar} = 39.9$$ u.

$$\frac{v_{Ar}}{490} = \sqrt{\frac{70.9}{39.9}}$$

Compute the ratio:

$$\frac{70.9}{39.9} = 1.7770$$

$$\sqrt{1.7770} = 1.3330$$

Therefore:

$$v_{Ar} = 490 \times 1.3330 = 653.2 \text{ m/s}$$

This is closest to $$651.7$$ m/s. Let me verify with more precision:

$$\frac{70.9}{39.9} = 1.77694...$$

$$\sqrt{1.77694} = 1.33302...$$

$$490 \times 1.33302 = 653.18$$

The closest option is $$651.7$$ m/s (Option B).

The answer is Option B: $$651.7$$ m/s$$^{-1}$$.

For the bimetallic strip to maintain a constant length difference (20 cm), both metals must expand by the same amount when the temperature changes.

The change in length for each material is:

$$\Delta L = L \alpha \Delta T$$

For the difference to remain constant, the expansions must be equal:

$$\Delta L_{\text{brass}} = \Delta L_{\text{iron}}$$

$$L_{\text{brass}} \times \alpha_{\text{brass}} \times \Delta T = L_{\text{iron}} \times \alpha_{\text{iron}} \times \Delta T$$

Cancelling $$\Delta T$$ from both sides:

$$L_{\text{brass}} \times \alpha_{\text{brass}} = L_{\text{iron}} \times \alpha_{\text{iron}}$$

$$40 \times 1.8 \times 10^{-5} = L_{\text{iron}} \times 1.2 \times 10^{-5}$$

$$L_{\text{iron}} = \dfrac{40 \times 1.8}{1.2} = \dfrac{72}{1.2} = 60 \text{ cm}$$

We can verify: The difference $$L_{\text{iron}} - L_{\text{brass}} = 60 - 40 = 20 \text{ cm}$$, which is the required scale length.

Therefore, the length of iron is $$\boxed{60}$$ cm.

We are told that the number density (molecules per unit volume) at a distance $$r$$ from the origin is $$n(r)=n_0 e^{-\alpha r^4}$$. To obtain the total number of molecules $$N$$ present in the whole space, we must integrate this density over the entire volume.

Because the density depends only on the radial distance, the gas is spherically symmetric. In spherical polar coordinates the volume element is $$dV = 4\pi r^2\,dr$$. Hence, using the definition $$N=\displaystyle\int_{\text{all space}} n(r)\,dV$$, we write

$$ N = \int_0^{\infty} n(r)\;4\pi r^2\,dr = 4\pi n_0\int_0^{\infty} r^2 e^{-\alpha r^4}\,dr. $$

Now we evaluate the integral $$I=\displaystyle\int_0^{\infty} r^2 e^{-\alpha r^4}\,dr.$$ To remove the exponential’s quartic argument, we perform the substitution $$x=\alpha r^4.$$ Then $$r = \left(\dfrac{x}{\alpha}\right)^{1/4}, \qquad dr = \dfrac{1}{4}\,\alpha^{-1/4}\,x^{-3/4}\,dx.$$

We also need $$r^2$$ in terms of $$x$$:

$$ r^2 = \left(\dfrac{x}{\alpha}\right)^{1/2} = x^{1/2}\,\alpha^{-1/2}. $$

Substituting $$r^2$$ and $$dr$$ into $$I$$ gives

$$ I = \int_0^{\infty} \Bigl(x^{1/2}\alpha^{-1/2}\Bigr)\, e^{-x}\, \Bigl(\frac{1}{4}\alpha^{-1/4}x^{-3/4}\Bigr)\,dx. $$

Collecting powers of $$x$$ and $$\alpha$$:

$$ x^{1/2}x^{-3/4}=x^{-1/4}, \qquad \alpha^{-1/2}\alpha^{-1/4}= \alpha^{-3/4}. $$

Thus

$$ I = \frac{1}{4}\,\alpha^{-3/4}\int_0^{\infty} x^{-1/4}e^{-x}\,dx. $$

The remaining integral is a standard Gamma-function. Recall the formula $$\displaystyle\Gamma(s)=\int_0^{\infty}x^{\,s-1}e^{-x}\,dx.$$ Here $$x^{-1/4}=x^{\,(-1/4)}=x^{\,\,(3/4-1)}$$, so $$s=\frac{3}{4}$$. Therefore

$$ \int_0^{\infty}x^{-1/4}e^{-x}\,dx = \Gamma\!\left(\frac{3}{4}\right), $$

a pure numerical constant, independent of $$\alpha$$ and $$n_0$$. Consequently,

$$ I = \frac{1}{4}\,\Gamma\!\left(\frac{3}{4}\right)\,\alpha^{-3/4}. $$

Returning to the expression for $$N$$, we substitute this result:

$$ N = 4\pi n_0\,I = 4\pi n_0\left[\frac{1}{4}\Gamma\!\left(\frac{3}{4}\right)\alpha^{-3/4}\right] = \pi\,\Gamma\!\left(\frac{3}{4}\right)\,n_0\,\alpha^{-3/4}. $$

The factors $$\pi$$ and $$\Gamma\!\left(\frac{3}{4}\right)$$ are fixed numerical constants, so we conclude that

$$ N \propto n_0\,\alpha^{-3/4}. $$

Hence, the correct answer is Option C.