InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Find the torque of a force `(7 hati + 3hatj - 5hatk)` about the origin. The force acts on a particle whose position vector is `(hati - hatj + hatk)`. |
|
Answer» Here,`oversetrarr(F) = 7hati + 3hatj - 5hatk, - overset rarr(r ) = hati - hatj + hatk` `overset rarr(tau) = overset rarr(r ) xx overset rarr(F) = (hati - hatj + hatk) xx (7hati + 3hatj - 5hatk)` `= |[hati,hatj,hatk],[1,-1,+1],[7,+3,-5]|` `oversetrarr(tau) = hati (5 - 3) - hatj (-5 - 7) + hatk (3 + 7)` `overset rarr(tau) = 2hati + 12 hatj + 10hatk` |
|
| 102. |
Two identical uniform discs roll without slipping on tow different sufaces AB and CD (see figure) starting at A and C with linear speeds `v_1 and v_2` respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed `v_1= 3m//s then v_2 in m//s is (g = 10 m//s^2)` |
|
Answer» Correct Answer - `(7)` Since final `KEs` are equal so `(KE "of rotation" + PE)_(A) = (KE "of rotation" + PE)_(C )` or `(1)/(2) [(3)/(2)mR^(2)] omega_(1)^(2) + mgh_(1) = (1)/(2) [(3)/(2) mR^(2)] omega_(2)^(2) + mgh_(2)` or `(3)/(4)v_(1)^(2) + gh_(1) = (3)/(4) v_(2)^(2) + gh_(2)` or `(3)/(4) (3)^(2) + 10 xx 30 = (3)/(4)v_(2)^(2) + 10 xx 27` On solving , `v^(2) = 49 or v = 7 ms^(-1)` |
|
| 103. |
Does the moment of inertia of abody change with the speed of rotation ? |
| Answer» No. Moment of inertia depends on mass of body, size of body, axis of rotation of the body and distribution of mass of the body about the axis of rotation. | |
| 104. |
What is the advantage of a flywheel ? |
| Answer» In a flywheel, the entire mass is distributed on the rim of the wheel at maximum distance from the axis of rotation. Therefore, moment of inertia of the flywheel is large. It helps in making the motion more smooth or less jerky. | |
| 105. |
Why spokes are provided in a bicycle wheel ? |
| Answer» By connecting the rim of the wheel to the axle through the spokes, the mass of wheel goes over to the rim. As a result, moment of inertia of the wheel increases. This ensures smooth motion. | |
| 106. |
A thin horizontal circular disc is roating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc.A. continuously decreasesB. continuously increasesC. first increases and then decreasesD. remain unchanged |
|
Answer» Correct Answer - C As external torque is zero, i.e. `tau = 0`, so angular momentum is conserved. `:. I omega = a` constant `= C` or `omega = (C )/(I) or omega prop (1)/(I)` When insect moves towards centre `I `decrease. Hence `omega` increases. As insect moves away from centre towards diametrically opposite point, `I` increases. Then `omega` decreases. Thus `omega` first increases and then decrease. |
|
| 107. |
What is moment of inertia of a solid sphere about its diameter ? |
| Answer» `I = (2)/(5) MR^(2)`, where `M` is the mass and `R` is radius of the solid sphere. | |
| 108. |
Find radius of gyration of a rod of length `l` and mass `m` about an axis perpendicular its length through one end. |
|
Answer» Here, `K = ?` According to theorem of parallel axes, `I_(end) = I_(cm) + m(l//2)^(2) = (ml^(2))/(12) + (ml^(2))/(4) = (ml^(2))/(3)` `:.mK^(2) = (ml^(2))/(3) or K = (l)/(sqrt(3))` |
|
| 109. |
Determine the coordinates of the centre of mass of a right circular solid cone of base radius `R` and height `h`. |
| Answer» It lies on the joining the apex to the centre of the base at a distance equal to `(1)/(4)th` of the length of this from the base. | |
| 110. |
What is nature of motion of cm of an isolated system ? |
| Answer» The `cm` of an isolated system remains at rest or move with a uniform velocity as no external force is acting on the isolated system. | |
| 111. |
What is an isolated system ? |
| Answer» An isolated system is that on which no external force is acting. | |
| 112. |
Name the rotational analogue of force. What are its units ? |
| Answer» Rotational analogue of force is torque. | |
| 113. |
Name the rotational analouge of force. What ar its units ? |
| Answer» Torque is rotational analogue of force. Its units are `N-m`. | |
| 114. |
Can centre of mass of a body coincide with the geometrical centre of the body ? |
| Answer» Yes, when a body has a uniform mass density, its centre of mass shall coincide with its geometrical centre. | |
| 115. |
A body A of mass M while falling wertically downwards under gravity brakes into two parts, a body B of mass `(1)/(3)` M and a body C of mass `(2)/(3)` M. The center of mass of bodies B and C taken together shifts compared to that of body A towards |
| Answer» The centre of mass of `B` and `C` does not shift compared to the centre of mass of `A`. It continues to move vertically downwards under gravity. This is because no new external force is involved in this case. | |
| 116. |
Moment of inertia of a body depends uponA. mass of bodyB. shape and size of bodyC. position and orientation of axis of rotationD. all the above |
| Answer» Correct Answer - D | |
| 117. |
A solid cylinder rolls up an inclined plane of angle of inclination `30^(@)`. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of `5 m//s`. (a) How far will the cylinder go up the plane ? (B) How long will it take to return to the bottom ? |
|
Answer» Here, `theta=30^(@), v=5 m//s` Let the cylinder go up the plane up to a height h. From `1//2 mv^(2)+1//2lW^(2)=mgh` `(1)/(2)mv^(2)+(1)/(2)((1)/(2)mr^(2))omega^(2)=mgh` `(3)/(4) mv^(2)=mgh` `h=(3v^(2))/(4g)=(3xx5^(2))/(4xx9.8)=1.913m` If s is the distance up the inclined plane, then as `sin theta =(h)/(s), s=(h)/(sin theta)=(1.913)/(sin 30^(@))3.856 m` Time taken to return to the bottom `t=sqrt((2s(1+(k^(2))/(r^(2))))/(g sin theta))=sqrt((2xx3.826(1+(1)/(2)))/(9.8 sin 30^(@)))=1.53s` |
|
| 118. |
A solid cylinder rolls down an inclined plane. Its mass is `2 kg and radius 0.1 m`. It the height of the inclined plane is `4m`, what is its rotational `K.E.` when it reaches foot of the plane ? Assume that the surfaces are smooth. Take `M.I.` of solid cylinder about its axis = `mr^(2)//2`. |
|
Answer» `K_(T) + K_(R) = K` `(1)/(2)mv^(2) + (1)/(2)I omega^(2) = K` But `(1)/(2)I omega^(2) = (1)/(2) ((1)/(2)mr^(2)) omega^(2) = (1)/(4)mv^(2)` `:. (1)/(2)mv^(2) + (1)/(4)mv^(2) = K or (3)/(4)mv^(2) = K` Rotational `KE, (1)/(2)I omega^(2) = (1)/(4)mv^(2) = (K)/(3)` `=(mgh)/(3) = (2 xx 9.8 xx 4)/(3) = 26 .13 J` |
|
| 119. |
A car moves on a road with a speed of `54 km h^(-1)`. The radius of its wheels is `0.35 m`. What is the average negative torque transmitted by its brakes to the wheels if the car is brought to rest in `15 s`? Moment of inertia of the wheels about the axis of rotation is `3 kg m^(2)`. |
|
Answer» Here, `u = 54 km h^(-1) = 15 ms^(-1)`, `r = 0.35 m` `tau = ?, t = 15 s, omega_(2) = 0 , omega_(1) = (u)/(r ) = (15)/(0.35) rad s^(-1)` From `omega_(2) - omega_(1) = alpha t` `alpha = (omega_(2) - omega_(1))/(t) = (0 - 15//0.35)/(15) = - (100)/(35) rad//s^(2)` `tau = I alpha = 3 (-(100)/(3)) = - 8.57 kg m^(2) s^(-2)` |
|
| 120. |
In a unifrom circular motion, which of the following remain constantA. speed and `omega`B. acceleration and angualr acc.C. time period and `omega`D. velocity and position vector |
|
Answer» Correct Answer - A::B::C Speed and angualr speed `omega`, are always consatnt n circular motion acceleration is consatnt . As `a = r alpha`, so angualr acceleration is also consatnt. |
|
| 121. |
The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it ?A. The forces may be acting radially from a point on the axis.B. The forces amay be acting on the axis of rotation.C. the forces may be acting parallel to the axis of rotation.D. The torque caused by some forces may be equal and opposite to that caused by other forces. |
|
Answer» Correct Answer - A::B::C::D When net external torque on a system of particles about an axis is zero, i.e.., `overset rarr(tau) = overset rarr(r ) xx overset rarr(F) = r F sin theta hat tau = Zero`, where, `theta` is angle between `overset rarr(r ) and overset rarr(F), hat tau` is unit vector along `overset rarr(tau)`, them all the four statements (a),(b),(c ),(d) are compatible. |
|
| 122. |
A preson is standing on the centre of a rotating table with his arms outstretched. The table is rotating freely with an angular his hands towards his chest. The moment of inertia reduces to `(3)/(5)` time the original moment of inertia. Calculate angular speed of the man when he withdraws his hands. |
|
Answer» Let `I_(1) = I, omega_(1) = 24` rpm `I_(2) = (3)/(5)I, omega_(2) = ?` According to the law of conservation of angular momentum, `I_(2)omega_(2) = I_(1)omega_(1)` `omega_(2) = (I_(1)omega_(1))/(I_(2)) = (I xx 24)/(3//5I) = 40`rpm |
|
| 123. |
Total linear momentum of a system of particles is equal to ……………..of the system and velocity of ……… . |
| Answer» product of total mass , Its centre of mass. | |
| 124. |
Torque and work are both equal to force time distance. How do they differ ? |
| Answer» Work is a scalar quanitiy, but torque is vector. Work = force xx distance moved by the body in the direction of force. Torque is measured by force multiplied by perpendicular distance of the line of action of force from the axis of the rotation. | |
| 125. |
Which physical quantities are expressed by the following : (i) rate of change of angular momentum (ii) moment of linear momentum ? |
|
Answer» (i) Rate of change of angular momentum represents torque. (ii) Moment of linear momentum represents angular momentum. |
|
| 126. |
Is a body in circular motion in equillibrium ? |
| Answer» No. This is because a body in circular motion has centripetal acceleration directed towards the centre of the circle. | |
| 127. |
If angular momentum is conserved in a system whose moment of inertia is decreased, will its rotational kinetic energy be also conserved ? Explain. |
|
Answer» Here, `L = I omega =` constant `K.E.` of rotation, `K = (1)/(2) I omega^(2)` `K = (1)/(2I)I^(2) omega^(2) = (L^(2))/(2I)` As `L` is constant, `:. K prop 1//l` When moment of inertia `(I)` decrease, `K.E.` of rotation `(K)` increase. Thus `K.E.` of rotation is not conserved. |
|
| 128. |
In circular motion, which of the following relations are valid ?A. `overset rarr(upsilon) = overset rarr(omega) xx overset rarr(r )`B. `overset rarr(a_(t)) = overset rarr(alpha) xx overset rarr(r )`C. `overset rarr(a_(t)) = overset rarr(omega) xx overset rarr(r )`D. `overset rarr(a) = overset rarr(omega) xx overset rarr(a_(t))` |
|
Answer» Correct Answer - A::B::C `overset rarr(upsilon) = overset rarr(omega) xx overset rarr(r ), overset rarr(a_(t)) = overset rarr(alpha) xx overset rarr_(r ), overset rarr(a_(r)) = overset rarr(omega) xx overset rarr_(r )` These are valid relations, but `overset rarr(alpha) = overset rarr(omega) xx overset rarr(a_(t)` is not valid. |
|
| 129. |
A solid cylinder and a hollow cylinder, both of the same mass and same external fiameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first ?A. Solid cylinderB. Both togetherC. One with higher densityD. Hollow cylinder |
|
Answer» Correct Answer - A Solid cylinder reaches bottom first because for solid cylinder, `(K^(2))/(R^(2)) = (1)/(2)` and for hollow cylinder, `(K^(2))/(R^(2)) = 1`. Acceleration down the inclined plane is Proportional to `(1)/(K^(2)//R^(2))`. Solid cylinder has greater acceleration. It reaches the bottom first. |
|
| 130. |
Two identical uniform rods of length `l` are joined to from `L` shaped frame, as shown is Fig. Locate the centre of mass of the frame. |
|
Answer» Let `m` be the mass of each rod. In Fig. `B` is taken as origin. `BC` is X-axis and `BA` is Y-axis. As the rods are uniform, their individual centre of mass must be lying at there respective centres, say `P` and `Q`. Thus the system is equivalent to two point masses `m` each, at `P (l//2,0)` and `Q(0,l//2)` `:. x_(cm) = (mxxl//2 + m xx0)/(m+m) = l//4` and `y_(cm) = (m xx 0 + m xx l//2)/(m + m) = l//4` Hence centre of mass of the system is at `O(l//4 , l//4)`. |
|
| 131. |
An autmobile moves on road with a speed of `54 km//h`. The radius of its wheel is `0.45 m` and the moment of inertia of the wheel about its axis of rotation is `3 kg m^(2)`. If the vehicle is brought to rest in `15 s`, the magnitude of average torque tansmitted by its brakes to the wheel is :A. `2.86 kg m^(2) s^(-2)`B. `6.66 kg m^(2) s^(-2)`C. `8.58 kg m^(2) s^(-2)`D. `10.86 kg m^(2) s^(-2)` |
|
Answer» Correct Answer - B Here, `u = 54 km//h = 15 m//s, r = 0.45m`, `I = 3kg m^(2), t = 15 sec, omega = 0` Now `u = r omega_(0) or omega_(0)= (u)/(r ) = (15)/(0.45) = (100)/(3)` As `omega = omega_(0) + alpha t` `0 = (100)/(3) + (-alpha) (15)` [As force is retarding] or `alpha = (100)/(54)`, Also `tau = I alpha` `= 3 xx (100)/(45) = 6.66 kg m^(2) s^(-2)` |
|
| 132. |
A rigid body is made of three indentical thin rods, each of length `L`, fastened together in the form of letter `H`, Fig. The body is free to rotate about a horizontal axis that tuns along the length of one of legs of `H`. The body is allowed to fall from rest from a position in which plane of `H` is horizontal. the angular speed of body when plane of `H` is vertical is A. `sqrt(g//L)`B. `(1)/(2) sqrt(g//L)`C. `2 sqrt(g//L)`D. `(2)/(3)sqrt(g//L)`s |
|
Answer» Correct Answer - D Moment of inertia of the system of three rods about the given axis, `I = 0 + (ml^(2))/(3) + mL^(2) = (4)/(3)mL^(2)` As decrease in gr. `P.E. =` increase in rot. `K.E.` `:. Mg xx (L)/(2) + mg (L) = (1)/(2)I omega^(2)` `(3)/(2)mgL = (1)/(2) ((4)/(3)mL^(2))omega^(2)` `omega = (3)/(2) sqrt((g)/(L))` |
|
| 133. |
Why is it easier to open a tap with two fingers than with onw finger ? |
| Answer» With two fingers, we are applying a couple whose arm is equal to length of handle of the tap. When we apply force with one finger, an equal and opposite force acts at the axis of rotation. Length of arm of couple is halved. Moment of couple is halved. | |
| 134. |
A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is A. always radially outwardsB. always radially inwardsC. radially outwards initially and radially inwards laterD. radially inwards initially and radially outwards later. |
|
Answer» Correct Answer - D When bead is at `A`, the force it applies on the wire is redially inwards. When it reaches at `B`, it strikes the surface and pushed outwards. Thus, force is radially outwards. |
|
| 135. |
If there is no external force acting on a nonrigid body, which of the followhng quantities must remain constant?A. linear momentumB. moment of inertiaC. angular momentumD. kinetic energy |
|
Answer» Correct Answer - A::C When no external force is acting, the linear momentum and angular momentum of the body remain constant. As the body is non rigid, its moment of inertia may change and hance its `K.E.` may also change. |
|
| 136. |
Show that angular momentum of a satellite of mass `M_(s)` revolving around earth of mass `M_(e)` in an orbit of radius `r` is `sqrt(GM_(s)^(2)M_(e)r)`. |
|
Answer» A satellite revolues around earth under gravitaitonal pull which provides the necessary centripetal force, i.e., `M_(S) = (upsilon^(2))/(r ) = (GM_(e)M_(s))/(r^(2))` `upsilon = sqrt((GM_(e))/(r ))` Angular momentum of satellite `L = M_(s). upsilon. r` `L = M_(s) sqrt((GM_(e))/(r )) xx r = sqrt(GM_(s)^(2)M_(e) r)` |
|
| 137. |
A man stands on a rotating platform, with his arms stretched horizontal holding a `5 kg` weight in each hand. The angular speed of the platform is `30` revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from `90 cm` to `20 cm`. moment of inertia of the man together with the platform may be taken to be constant and equal t `7.6 kg m^(2)`. (a) What is the his new angular speed ? (Neglect friction.) (b) Is kinetic energy conserved in the process ? If not, from where does the change come about ? |
|
Answer» Here, `l_(1)=7.6+2xx5(0.9)^(2)=15.7 kg m^(2)` `w_(1)=30 rpm` `l_(2)=7.6+2xx5(0.2)2=8.0 kg m^(2)` `w_(2)=?` According to the principle of conservation of angular momentum, `l_(2)w_(2)=l_(1)w_(1)` No, kinetic energy is not conserved in the process. In fact, as moment of inertia decreases, K.E. of rotation increases. This change comes about as work is done by the man in bringing his arms closer to his body. |
|
| 138. |
Point masses `m_(1) and m_(2)` are placed at the opposite ends of a rigid rod of length `L`, and negligible mass. The rod is to be set rotating about an axis perpendicualr to it. The position of point `P` on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity `omega_(0)` is minimum, is given by : A. `x = (m_(2)L)/(m_(1) + m_(2))`B. `x = (m_(1)L)/(m_(1) + m_(2))`C. `x = (m_(1))/(m_(2))L`D. `x = (m_(2))/(m_(1))L` |
|
Answer» Correct Answer - A As `K.E.` of rotation `= (1)/(2)I omega^(2)` `I` is min. about the centre of mass `:. M_(1)x = m_(2) (L - x) or x = (m_(2)L)/(m_(1) + m_(2))` |
|
| 139. |
Assertion : There are two propellers in a helicopter. Reason : Angular momentum is conserved.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
|
Answer» Correct Answer - B Both, the assertion and reason are true, but the reason is not a correct explanation of the assertion. Infact, if helicopter has only one propeller, the helicopter itself will turn in opposite direction to conserve the angular momentum. |
|
| 140. |
When no external torque acts on a system, its angular momentum remains constant. Is the statement true ? Should kinetic energ of rotation of the system remain constant ? If yes, why , and if no, why not? |
|
Answer» Yes, the statement is true. This is as per the principle of conservation of angular momentum, i.e. When `tau = 0, L = I omega =` constant. Now, `K.E`. of rotation `= (1)/(2)I omega^(2)` `E = (I^(2)omega^(2))/(2I) = (L^(2))/(2I)` As `L` is constant, therefore, `E prop (1)/(I)` i.e., `K.E.` of rotation of the system shall NOT remain constant. If moment of inertia `//` of the system increase, `K.E` of rotation `E` will decrease and vice-verse. However, K.E. of rotation will remain constant if moment `//` of the body is maintained constant. |
|
| 141. |
A torque of `10^(3) N-m` acting on a rigid body, turns it throughr `30^(@)` in `0.2` second. Calculate work done by the body and power of torque. |
|
Answer» Here, `tau = 10^(3) N-m` `theta = 30^(@) = (pi)/(6) radian, t = 0.2 s`. `W = tau theta = 10^(3) xx (tau)/(6) = (22)/(7 xx 6) xx 10^(3)J` `= 5.23 xx 10^(2)J` `p = tau omega = (W)/(t) = (5.23 xx 10^(2))/(0.2) = 2.615 xx 10^(3) watt` |
|
| 142. |
Is moment of inertia a scalar or a vector ? |
| Answer» Moment of inertia is a scaler quantity. | |
| 143. |
A body is in rotational motion. Is it necessary that a torque be acting on it ? |
| Answer» No. torque is required only for producing angular acceleration. | |
| 144. |
What is the dimensional formula of angular momentum and what are its units ? Is it a scalar ? |
| Answer» The dimensional formula of angular momentum is `[M^(1)L^(2)T^(-1)]`. Its units are `kg m^(2) s^(-1)`. It is not a scalar, but a vactor. | |
| 145. |
A car weighs `1800 kg`. The distance between its front and back axles is `1.8 m`. Its centre of gravity is `1.05 m` behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. |
|
Answer» Let `F_(1) " and " F_(2)` be the forces exerted by the level ground on front wheels and back wheels respectively. Considering rotational equilibrium about the front wheels, `F_(2) xx1.8= mgxx1.05 " or " F_(2)=1.05//1.8xx1800xx9.8 N=10290 N` Force on each back wheel is `10290//2` N or 5145 N. Considering rotational equilibrium about the back wheels. `F_(1)xx1.8=mg(1.8-1.05)=0.75xx1800xx9.8 " or " F_(1)=0.75xx1800xx9.8//1.8=7350 N` Force on each front wheel is `7350//2 N` or 3675 N. |
|
| 146. |
A non-uniform bar of weight `W` is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are `36.9^(@) and 53.1^(@)` respectively. The bar is `2m` long. Calculate the distance `d` of the centre of gravity of the bar from its left end. |
|
Answer» As it is clear from Fig., `theta_(1)=36.9^(@), theta_(2)=53.1^(@)`. It `T_(1),T_(2)` are the tensions in the two strings, then for equilibrium along the horizontal, `T_(1) sin theta_(1)=T_(2) sin theta_(2)` or ` (T_(1))/(T_(2))=(sin theta_(2))/(sin theta_(1))=(sin 53.1^(@))/(sin36.9^(@))` `=(0.7404)/(0.5477)=1.3523` Let d be the distance of centre of gravity C of the bar from the left end. For rotaional equilibrium about C, `T_(1) cos theta_(1) xxd=T_(2) cos theta_(2) (2-d)` `T_(1) cos 36.9^(@)xxd=T_(2) cos 53.1^(@) (2-d)` `T_(1)xx0.8366 d= T_(2)xx0.6718 (2-d)` Put `T_(1)=13523 T_(2)` and solve to get d=0.745 m `tau=I_(1)alpha_(1)=I_(2)alpha_(2)` or ` (alpha_(1))/(alpha_(2))=(I_(2))/(I_(1))=((2//5)MR^(2))/(MR^(2))=(2)/(5)` or ` alpha_(2)=(5)/(2) alpha_(1)`. |
|
| 147. |
A fire cracker following a parabolic path explodes in mid air. The centre of mass of all the fragments will follow a pathA. along horizontalB. along verticalC. along same parabolaD. along circulae |
| Answer» Correct Answer - C | |
| 148. |
The motor of an engine is rotating about its axis with an angular velocity of `100`rpm. It comes to rest in `15s`, after being switched off. If angular deceleration is constant, what is the number to revolutions made by it before coming to rest ? |
|
Answer» Here, `n_(1) = 100"rpm" = (100)/(60) rps`, `n_(2) = 0, t = 15 s, alpha = ?, n = ?` `alpha = (omega_(2) - omega_(1))/(t) = (2pi(n_(2) - n_(1)))/(t)` `= (2pi(0-(100)/(60)))/(15) = - (2pi)/(9) rad s^(-2)` `theta = omega_(1)t + (1)/(2) alpha t^(2)` `= 2pi xx (100)/(60) xx 15 + (1)/(2) (-(2pi)/(9)) (15)^(2)` `= 50pi - 25 pi = 25 pi` Number of rotations made, `n = (theta)/(2pi) = (25pi)/(2pi) = 12.5` |
|
| 149. |
Angle traced by a rotating body in nth seconds is `theta_(nth)=`…………where symbols have their usual meaning. |
| Answer» `theta_(nth) = omega_(1) + (alpha)/(2) (2n - 1)` | |
| 150. |
What is moment of inertia of a solid cylinder of mass `m`, length l and radius` r` about the axis of the cylinder ? |
| Answer» Correct Answer - `I = (1)/(2)mr^(2)` | |