Question

In List -I a constant force F is applied on the rod of mass m and length 'l' such that in each case rod moves. A transverse pulse is created at the end point P in each case. The time to move the pulse from P to Q is given in List -II. Then match the following.

• A. (m, l) P →F μ = 0 (m = 4kg, l = 20 m, F = 5N)
• B. (m, l) P Q →F μ = 1/20 (m = 9kg, l = 32 m, F = 8N)
• C. ↑F Q (m, l) P (m = 7kg, l = 14 m, F = 2N)
• D. F (m, ) μ= 1/10 P 30° (m = 10kg, l = 20 m, F = 2N)

Answer 1

I->R, II->P, III->S, IV->Q

Answer 2

The speed of a transverse wave on a rod under tension $T$ and linear mass density $\mu$ is given by $v = \sqrt{\frac{T}{\mu}}$. The linear mass density of the rod is $\mu = \frac{m}{l}$. The time taken for the pulse to travel a distance $l$ is $t = \int_0^l \frac{dx}{v(x)}$, where $x$ is the distance from point P.

Case (I): Horizontal rod on a frictionless surface. Force $F$ is applied at Q, pulse starts at P. The acceleration of the rod is $a = \frac{F}{m}$. Consider a segment of the rod from P to a point at distance $x$. The mass of this segment is $m' = \frac{m}{l}x$. The tension $T(x)$ at distance $x$ from P is the force required to accelerate this segment. $T(x) = m'a = \frac{m}{l}x \frac{F}{m} = \frac{Fx}{l}$. The speed of the pulse at distance $x$ from P is $v(x) = \sqrt{\frac{T(x)}{\mu}} = \sqrt{\frac{Fx/l}{m/l}} = \sqrt{\frac{Fx}{m}}$. The time taken to travel from P to Q is $t = \int_0^l \frac{dx}{v(x)} = \int_0^l \frac{dx}{\sqrt{Fx/m}} = \sqrt{\frac{m}{F}} \int_0^l x^{-1/2} dx = \sqrt{\frac{m}{F}} [2x^{1/2}]_0^l = 2\sqrt{\frac{ml}{F}}$. Given $m=4$ kg, $l=20$ m, $F=5$ N. $t = 2\sqrt{\frac{4 \times 20}{5}} = 2\sqrt{16} = 2 \times 4 = 8$ seconds. (I) matches with (R).

Case (II): Horizontal rod on a surface with friction. Force $F$ is applied at Q, pulse starts at P. The acceleration of the rod is $a = \frac{F - \mu mg}{m}$. Consider a segment from P to $x$. Mass is $m' = \frac{m}{l}x$. Friction on this segment is $f'(x) = \mu m'g = \mu \frac{m}{l}x g$. The net force on this segment is $T(x) - f'(x) = m'a$. $T(x) = m'a + f'(x) = \frac{m}{l}x a + \mu \frac{m}{l}x g = \frac{m}{l}x (a + \mu g) = \frac{m}{l}x (\frac{F - \mu mg}{m} + \mu g) = \frac{m}{l}x (\frac{F - \mu mg + \mu mg}{m}) = \frac{m}{l}x \frac{F}{m} = \frac{Fx}{l}$. The tension distribution is the same as in Case (I). The time taken is $t = 2\sqrt{\frac{ml}{F}}$. Given $m=9$ kg, $l=32$ m, $F=8$ N. $t = 2\sqrt{\frac{9 \times 32}{8}} = 2\sqrt{9 \times 4} = 2\sqrt{36} = 2 \times 6 = 12$ seconds. (II) matches with (P).

Case (III): Vertical rod. Force $F$ is applied upwards at Q, pulse starts at P. The acceleration of the rod is $a = \frac{F - mg}{m}$. Consider a segment from P to $x$. Mass is $m' = \frac{m}{l}x$. Gravity on this segment is $m'g = \frac{m}{l}x g$. The net force on this segment is $T(x) - m'g = m'a$. $T(x) = m'g + m'a = \frac{m}{l}x g + \frac{m}{l}x a = \frac{m}{l}x (g + a) = \frac{m}{l}x (g + \frac{F - mg}{m}) = \frac{m}{l}x (\frac{mg + F - mg}{m}) = \frac{m}{l}x \frac{F}{m} = \frac{Fx}{l}$. The tension distribution is the same as in Case (I). The time taken is $t = 2\sqrt{\frac{ml}{F}}$. Given $m=7$ kg, $l=14$ m, $F=2$ N. $t = 2\sqrt{\frac{7 \times 14}{2}} = 2\sqrt{7 \times 7} = 2\sqrt{49} = 2 \times 7 = 14$ seconds. (III) matches with (S).

Case (IV): Rod on an inclined plane at 30 degrees. Force $F$ is applied horizontally at Q. The pulse starts at P. The rod moves.

The forces acting on the rod are: applied force $F$ (horizontal), gravity $mg$ (vertical), normal force $N$ (perpendicular to the plane), friction force $f_k = \mu N$ (up the plane). The rod is moving along the inclined plane. Let's resolve the forces along the plane and perpendicular to the plane. The angle of inclination is 30 degrees. Components of gravity: $mg \sin 30^\circ$ down the plane, $mg \cos 30^\circ$ perpendicular to the plane. Component of applied force $F$: $F \cos 30^\circ$ up the plane, $F \sin 30^\circ$ perpendicular to the plane, pointing into the plane. The normal force is $N - mg \cos 30^\circ - F \sin 30^\circ = 0$, so $N = mg \cos 30^\circ + F \sin 30^\circ$. The friction force is $f_k = \mu N = \mu (mg \cos 30^\circ + F \sin 30^\circ)$, acting up the plane. The net force along the plane is $F_{net} = F \cos 30^\circ - mg \sin 30^\circ - f_k = F \cos 30^\circ - mg \sin 30^\circ - \mu (mg \cos 30^\circ + F \sin 30^\circ)$. The acceleration of the rod along the plane is $a = \frac{F_{net}}{m}$. Consider a segment from P to a point at distance $x$ along the plane. Mass is $m' = \frac{m}{l}x$. Gravity on this segment along the plane is $m'g \sin 30^\circ$. Applied force component on this segment along the plane is $\frac{m'}{m} F \cos 30^\circ = \frac{x}{l} F \cos 30^\circ$. (Assuming the applied force is distributed along the length, which is not explicitly stated, but consistent with the tension method). Friction on this segment is $f'_k(x) = \mu (m'g \cos 30^\circ + \frac{m'}{m} F \sin 30^\circ) = \mu (\frac{m}{l}x g \cos 30^\circ + \frac{x}{l} F \sin 30^\circ) = \frac{x}{l} \mu (mg \cos 30^\circ + F \sin 30^\circ)$. The net force on this segment along the plane is $T(x) + \frac{x}{l} F \cos 30^\circ - m'g \sin 30^\circ - f'_k(x) = m'a$. $T(x) = m'a + m'g \sin 30^\circ + f'_k(x) - \frac{x}{l} F \cos 30^\circ$ $T(x) = \frac{m}{l}x a + \frac{m}{l}x g \sin 30^\circ + \frac{x}{l} \mu (mg \cos 30^\circ + F \sin 30^\circ) - \frac{x}{l} F \cos 30^\circ$ $T(x) = \frac{x}{l} [ma + mg \sin 30^\circ + \mu (mg \cos 30^\circ + F \sin 30^\circ) - F \cos 30^\circ]$ Substitute $ma = F \cos 30^\circ - mg \sin 30^\circ - \mu (mg \cos 30^\circ + F \sin 30^\circ)$. $T(x) = \frac{x}{l} [ (F \cos 30^\circ - mg \sin 30^\circ - \mu (mg \cos 30^\circ + F \sin 30^\circ)) + mg \sin 30^\circ + \mu (mg \cos 30^\circ + F \sin 30^\circ) - F \cos 30^\circ ]$ $T(x) = \frac{x}{l} [ F \cos 30^\circ - mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ + mg \sin 30^\circ + \mu mg \cos 30^\circ + \mu F \sin 30^\circ - F \cos 30^\circ ]$ $T(x) = \frac{x}{l} [0] = 0$. This result is incorrect. The tension should vary along the rod to cause varying acceleration of segments. Let's consider the tension at a point $x$ from P. The tension $T(x)$ is the force exerted by the part of the rod from $x$ to Q on the part from P to $x$. Consider the segment from $x$ to Q. The length is $l-x$, mass is $\frac{m}{l}(l-x)$. Forces on this segment along the plane: $F \cos 30^\circ$ (if applied at Q), $T(x)$ (down the plane at $x$), gravity component $\frac{m}{l}(l-x)g \sin 30^\circ$ (down the plane), friction $\mu (\frac{m}{l}(l-x)g \cos 30^\circ + \frac{l-x}{l} F \sin 30^\circ)$ (up the plane). Equation of motion: $F \cos 30^\circ - T(x) - \frac{m}{l}(l-x)g \sin 30^\circ - \mu (\frac{m}{l}(l-x)g \cos 30^\circ + \frac{l-x}{l} F \sin 30^\circ) = \frac{m}{l}(l-x)a$. $T(x) = F \cos 30^\circ - \frac{m}{l}(l-x)g \sin 30^\circ - \mu (\frac{m}{l}(l-x)g \cos 30^\circ + \frac{l-x}{l} F \sin 30^\circ) - \frac{m}{l}(l-x)a$. $T(x) = F \cos 30^\circ - \frac{l-x}{l} [mg \sin 30^\circ + \mu (mg \cos 30^\circ + F \sin 30^\circ) + ma]$. Substitute $ma = F \cos 30^\circ - mg \sin 30^\circ - \mu (mg \cos 30^\circ + F \sin 30^\circ)$. $T(x) = F \cos 30^\circ - \frac{l-x}{l} [mg \sin 30^\circ + \mu mg \cos 30^\circ + \mu F \sin 30^\circ + F \cos 30^\circ - mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ]$. $T(x) = F \cos 30^\circ - \frac{l-x}{l} [F \cos 30^\circ] = F \cos 30^\circ (1 - \frac{l-x}{l}) = F \cos 30^\circ \frac{x}{l}$. The tension at distance $x$ from P is $T(x) = \frac{Fx \cos 30^\circ}{l}$. The speed of the pulse at distance $x$ from P is $v(x) = \sqrt{\frac{T(x)}{\mu_{linear}}} = \sqrt{\frac{Fx \cos 30^\circ / l}{m/l}} = \sqrt{\frac{Fx \cos 30^\circ}{m}}$. The time taken to travel from P to Q is $t = \int_0^l \frac{dx}{v(x)} = \int_0^l \frac{dx}{\sqrt{Fx \cos 30^\circ/m}} = \sqrt{\frac{m}{F \cos 30^\circ}} \int_0^l x^{-1/2} dx = \sqrt{\frac{m}{F \cos 30^\circ}} [2x^{1/2}]_0^l = 2\sqrt{\frac{ml}{F \cos 30^\circ}}$. Given $m=10$ kg, $l=20$ m, $F=2$ N, $\mu=1/10$, $30^\circ$. $\cos 30^\circ = \frac{\sqrt{3}}{2}$. $t = 2\sqrt{\frac{10 \times 20}{2 \times \sqrt{3}/2}} = 2\sqrt{\frac{200}{\sqrt{3}}} = 2 \times 10\sqrt{\frac{2}{\sqrt{3}}} = 20 \sqrt{\frac{2}{\sqrt{3}}} \approx 20 \sqrt{\frac{2}{1.732}} \approx 20 \sqrt{1.154} \approx 20 \times 1.074 \approx 21.48$. This does not match any of the options in List-II.

Let's re-examine the problem statement and the diagram for Case (IV). The force F is applied horizontally at Q. The pulse is created at P. Let's assume $g=10$ m/s$^2$. In Case (IV), $m=10$ kg, $l=20$ m, $F=2$ N, $\mu=1/10$. $mg \sin 30^\circ = 10 \times 10 \times 0.5 = 50$ N. $F \cos 30^\circ = 2 \times \sqrt{3}/2 = \sqrt{3} \approx 1.732$ N. $mg \cos 30^\circ = 10 \times 10 \times \sqrt{3}/2 = 50\sqrt{3} \approx 86.6$ N. $F \sin 30^\circ = 2 \times 0.5 = 1$ N. Normal force $N = mg \cos 30^\circ + F \sin 30^\circ = 50\sqrt{3} + 1 \approx 86.6 + 1 = 87.6$ N. Maximum static friction $f_{s,max} = \mu_s N$. Assuming $\mu_s = \mu_k = \mu = 1/10$. $f_{k} = \mu N = \frac{1}{10} (50\sqrt{3} + 1) = 5\sqrt{3} + 0.1 \approx 8.66 + 0.1 = 8.76$ N. Force component up the plane due to F is $F \cos 30^\circ = \sqrt{3} \approx 1.732$ N. Force component down the plane due to gravity is $mg \sin 30^\circ = 50$ N. Net force along the plane (ignoring friction for now) is $F \cos 30^\circ - mg \sin 30^\circ = 1.732 - 50 = -48.268$ N. This means the net force is down the plane. The friction force opposes the motion. If the rod moves down the plane, friction is up the plane. The total force down the plane is $mg \sin 30^\circ = 50$ N. The total force up the plane is $F \cos 30^\circ + f_k = \sqrt{3} + \mu N = \sqrt{3} + \frac{1}{10} (50\sqrt{3} + 1) = \sqrt{3} + 5\sqrt{3} + 0.1 = 6\sqrt{3} + 0.1 \approx 10.39 + 0.1 = 10.49$ N. Since $50 > 10.49$, the rod will move down the plane. The net force down the plane is $F_{net} = mg \sin 30^\circ - F \cos 30^\circ - f_k = 50 - \sqrt{3} - (5\sqrt{3} + 0.1) = 50 - 6\sqrt{3} - 0.1 = 49.9 - 6\sqrt{3} \approx 49.9 - 10.39 = 39.51$ N. The acceleration down the plane is $a = \frac{F_{net}}{m} = \frac{39.51}{10} = 3.951$ m/s$^2$.

Let's recheck the tension calculation. Consider a point at distance $x$ from P. The segment from P to $x$ has mass $m' = \frac{m}{l}x$. Forces on this segment along the plane: tension $T(x)$ (up the plane at $x$), gravity component $m'g \sin 30^\circ$ (down the plane), friction $f'_k(x) = \mu (m'g \cos 30^\circ + \frac{x}{l} F \sin 30^\circ)$ (up the plane, since motion is down). Equation of motion down the plane: $m'g \sin 30^\circ - T(x) - f'_k(x) = m'a$. $T(x) = m'g \sin 30^\circ - f'_k(x) - m'a = \frac{m}{l}x g \sin 30^\circ - \mu (\frac{m}{l}x g \cos 30^\circ + \frac{x}{l} F \sin 30^\circ) - \frac{m}{l}x a$. $T(x) = \frac{x}{l} [mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ - ma]$. Substitute $ma = mg \sin 30^\circ - F \cos 30^\circ - \mu (mg \cos 30^\circ + F \sin 30^\circ)$. $T(x) = \frac{x}{l} [mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ - (mg \sin 30^\circ - F \cos 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ)]$. $T(x) = \frac{x}{l} [mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ - mg \sin 30^\circ + F \cos 30^\circ + \mu mg \cos 30^\circ + \mu F \sin 30^\circ]$. $T(x) = \frac{x}{l} [F \cos 30^\circ]$. This is the same tension distribution as before, $T(x) = \frac{Fx \cos 30^\circ}{l}$. The time taken is $t = 2\sqrt{\frac{ml}{F \cos 30^\circ}}$. Using $m=10$, $l=20$, $F=2$, $\cos 30^\circ = \sqrt{3}/2$. $t = 2\sqrt{\frac{10 \times 20}{2 \times \sqrt{3}/2}} = 2\sqrt{\frac{200}{\sqrt{3}}} = 20\sqrt{\frac{2}{\sqrt{3}}} \approx 21.48$ sec. This still does not match. Let's recheck the options.

Let's re-examine the tension calculation. Consider a point at distance $x$ from P. The segment from P to $x$ has mass $m' = \frac{m}{l}x$. The forces acting on this segment along the plane are: tension $T(x)$ (up the plane), gravity component $m'g \sin 30^\circ$ (down the plane), friction $f'_k(x)$ (up the plane, since motion is down). The equation of motion down the plane for this segment is $m'g \sin 30^\circ - T(x) - f'_k(x) = m'a$. $T(x) = m'g \sin 30^\circ - f'_k(x) - m'a$. $f'_k(x) = \mu N'(x)$. The normal force on the segment from P to $x$ is $N'(x) = m'g \cos 30^\circ + \frac{x}{l} F \sin 30^\circ$. $T(x) = \frac{m}{l}x g \sin 30^\circ - \mu (\frac{m}{l}x g \cos 30^\circ + \frac{x}{l} F \sin 30^\circ) - \frac{m}{l}x a$. $T(x) = \frac{x}{l} [mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ - ma]$. Let $C = mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ$. This is the net force per unit mass component along the plane due to gravity and friction. $T(x) = \frac{x}{l} [C - ma]$. This still doesn't look right.

Let's consider the forces acting on the segment from $x$ to Q. Length $l-x$, mass $\frac{m}{l}(l-x)$. Forces along the plane: $F \cos 30^\circ$ (up at Q), $T(x)$ (down at $x$), gravity component $\frac{m}{l}(l-x)g \sin 30^\circ$ (down), friction $\mu (\frac{m}{l}(l-x)g \cos 30^\circ + \frac{l-x}{l} F \sin 30^\circ)$ (up, since motion is down). Equation of motion down the plane: $\frac{m}{l}(l-x)g \sin 30^\circ + T(x) - F \cos 30^\circ + \mu (\frac{m}{l}(l-x)g \cos 30^\circ + \frac{l-x}{l} F \sin 30^\circ) = \frac{m}{l}(l-x)a$. $T(x) = \frac{m}{l}(l-x)a - \frac{m}{l}(l-x)g \sin 30^\circ + F \cos 30^\circ - \mu (\frac{m}{l}(l-x)g \cos 30^\circ + \frac{l-x}{l} F \sin 30^\circ)$. $T(x) = \frac{l-x}{l} [ma - mg \sin 30^\circ + \mu mg \cos 30^\circ + \mu F \sin 30^\circ] + F \cos 30^\circ$. $ma = mg \sin 30^\circ - F \cos 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ$. $T(x) = \frac{l-x}{l} [mg \sin 30^\circ - F \cos 30^\circ - \mu mg \ cos 30^\circ - \mu F \sin 30^\circ - mg \sin 30^\circ - \mu mg \cos 30^\circ - \mu F \sin 30^\circ] + F \cos 30^\circ$. $T(x) = \frac{l-x}{l} [-F \cos 30^\circ - 2\mu mg \cos 30^\circ - 2\mu F \sin 30^\circ] + F \cos 30^\circ$. This also does not look like a simple form.

Let's assume the tension calculation $T(x) = \frac{Fx \cos 30^\circ}{l}$ is correct, meaning the effective force causing tension variation is the component of F along the rod. Then $t = 2\sqrt{\frac{ml}{F \cos 30^\circ}}$. $m=10, l=20, F=2, \cos 30^\circ = \sqrt{3}/2$. $t = 2\sqrt{\frac{10 \times 20}{2 \times \sqrt{3}/2}} = 2\sqrt{\frac{200}{\sqrt{3}}} = 20\sqrt{\frac{2}{\sqrt{3}}}$. If we use $g=10$, then $\sqrt{3} \approx 1.732$. $t \approx 21.49$. Let's assume the intended answer for Case (IV) is 20 sec. If $t = 20$, then $20 = 2\sqrt{\frac{10 \times 20}{2 \cos 30^\circ}} = 2\sqrt{\frac{100}{\cos 30^\circ}}$. $10 = \sqrt{\frac{100}{\cos 30^\circ}}$. $100 = \frac{100}{\cos 30^\circ}$. This implies $\cos 30^\circ = 1$, which is false.

There seems to be a mismatch in Case (IV). However, given that the first three cases match perfectly, it is likely that the formula for time is correct, and there might be a slight discrepancy in the given values or the options for Case (IV). Since 21.49 is closest to 20, let's assume (IV) matches with (Q).

Final matches: (I) - (R) 8 sec (II) - (P) 12 sec (III) - (S) 14 sec (IV) - (Q) 20 sec (based on closest value)