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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.
| 151. |
The moment of inertia of a uniform cylinder of length `l and radius R` about its perpendicular bisector is `I`. What is the ratio `l//R` such that the moment of inertia is minimum ?A. `1`B. `(3)/(sqrt(2))`C. `sqrt((3)/(2))`D. `(sqrt(3))/(2)` |
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Answer» Correct Answer - C Moment of inertia of a uniform cylinder of length `l and radius R` about its perpendicular bisector is `I = (ml^(2))/(12) + (mR^(2))/(4) = (m)/(4)((l^(2))/(3)+R^(2))` ..(i) and `m = pi R^(2)l.rho` `:. R^(2) = (m)/(pi//rho)` From(i), `I = (m)/(4) ((l^(2))/(3)+(m)/(pi lrho))` `I` will be minimum, when `(dI)/(dl) = 0` `(d)/(dl) [(m)/(4) ((l^(2))/(3)+(m)/(pi lrho))] = 0` `(m)/(4) ((2l^(2))/(3)+(m)/(pil^(2)rho)) = 0` or `(2l)/(3) = (m)/(pil^(2)rho) = (piR^(2)l rho)/(pi l^(2)rho) = (R^(2))/(l)` or `(l^(2))/(R^(2)) = (3)/(2) , (l)/(R ) = sqrt((3)/(2))` |
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| 152. |
About which axis would the moment of inertia of a body be minimum ? |
| Answer» The moment of inertia of a body is minimum about an axis passing through its centre of mass. | |
| 153. |
A solid wooden sphere rolls down two different inclined planes of the same height but of different inclinations. (a) Will it reach the bottom with same speed in each case ? (b) Will it take longer to roll down one inclined plane than other ? Explain. |
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Answer» (a) Yes, because speed at the bottom depends only on height and not on slope. (b) Yes, greater the inclination `(theta)`, smaller will be the time of descent, as `t prop 1//sin theta`. |
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| 154. |
A circular plate of unifrom thickness has a diameter of `56 cm`. A circular portion of diameter `42 cm` is removed from the edge of the plate as shown in Fig. Find the position of centre of mass of the remaining portion. |
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Answer» Suppose mass per unit area of circular plate `= m`. `:.` Total mass of circular plate, `M = m xx pi(d//2)^(2) = pi m((56)/(2))^(2) = 784 pi m` This is supposed to be concentrated at the centre `O` of the disc. Mass of cut portion, `M_(1) = m xx pi ((d_(1))/(2))^(2) = pi m(42//2)^(2) = 441 pi m` This is supposed to be concentrated at `O_(1)`, where `O_(1) O = AO - AO_(1) = 28 - 21 = 7 cm`. Mass of remaining portion of disc, `M_(2) = M - M_(1) = 784 pi m - 441pi m = 343 pi m` Let it be concentrated at `O_(2)`, where `OO_(2) = x`. As `x_(cm) = (M_1)x_(1) + (M_(2)x_(2))/(M_(1) + M_(2))` and `x_(cm) = 0 (at O) :. M_(1)x_(1) + M_(2)x_(2) = 0` `441pi m xx 7+ 343 pi m x = 0` or `x = (-441pi m xx 7)/(343 pi m) =- 9cm` Hence, centre of mass of remaining portion of disc is at `9 cm` from centre of disc on the right side of `O`. |
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| 155. |
A unifrom square plate `S (side c)` and a unifrom rectangular plate `R(side b,a)` have identical areas and mass [Fig.] Show that (i) `I_(xR)//I_(xS) lt 1`, (ii) `I_(yR)//I_(yS) gt 1`, (iii) `I_(zR)//I_(zS) gt 1`. |
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Answer» As area of square plate `S = area` of rectangular plate `R` `c^(2) = a xx b` Now, by definition (i) `(I_(xR))/(I_(xS)) = (b^(2))/(c^(2))` As `b lt c, (I_(xR))/(I_(xS)) lt 1` (ii) `(I_(yR))/(I_(yS)) = (a^(2))/(c^(2))` As `a gt c, (I_(yR))/(I_(yS)) gt 1`. (iii) `(I_(zR) - I_(zS)) prop (a^(2) + b^(2) - 2c^(2))` `prop (a^(2) + b^(2) - 2 ab)` `prop (a - b)^(2)`, which is + `:. (I_(xR) I_(xS)) gt 0 or (I_(xR))/(I_(xS)) gt 1` |
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| 156. |
If polar ice caps melt, then the time duration of one day |
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Answer» Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concentrated near the axis of rotation spreads out. Therefore, As no external torque axts, `:. L = I omega = I ((2pi)/(T)) =` constant. With increase of `I, T` will increase i.e., lenght of the day will increase. |
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| 157. |
A raw egg and a hard boiled egg are made to spin on a table with the same angular speed about the same axis. The ration of the time taken by the two to stop isA. `= 1`B. `lt 1`C. `gt 1`D. none of the above |
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Answer» Correct Answer - B A raw egg has some fluid in it and a hard boiled egg is solid from inside. On spinning, the fluid is thrown outwards. Therefore, `I_(r) gt I_(b)` As `I xx omega = constant :. omega_(r) lt omega_(b)` or `(omega_(r))/(omega_(b)) lt 1 :. (t_(r))/(t_(b)) lt 1` |
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| 158. |
Why spin angular velocity of a star is greatly enhanced when it collapses under gravitational pull and becomes a neutron star ? |
| Answer» On collapsing under gravitational pull, size of star decreases. Therefore, its moment of inertia decreases. As angular momentum `(L = I omega)` is conserved, and `I` decreases, therefore, spin angular velocity `omega` increases. | |
| 159. |
For which of the following does the centre of mass lie outside the body ?A. A pencilB. A shotputC. A diceD. A bangle |
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Answer» Correct Answer - D Centre of mass of a bangle lies at the centre of the bangle, which is outside the body. |
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| 160. |
Can centre of mass of a body lie where there is absolutely no mass ? |
| Answer» Yes, it can. For example, centre of mass of a uniform circular ring lies at the centre of the ring, where there is no mass. | |
| 161. |
Where does the centre of mass of a uniform rectangular lamina lie ? |
| Answer» `CM` lies at the point of intersection of diagonals of the rectangle. | |
| 162. |
The torque `(tau)` due to a force gives us the turning effect of the force about a fixed point/axis. It is measured by the product of magnitude of force `(F)` and perpendicular distance `(r )` of the line of action of force form the axis of rotation. `overset rarr(tau) = overset rarr(r ) xx overset rarr(F) = rF sin theta hatn` where `thete` is smaller angle between `overset rarr(r )` and `hatn` is unit vector along `overset rarr(tau)`. Read the above passage and answer the following questions : (i) What is the significance of torque ? (ii) How do you determine the direction of torque ? (ii) What is the implication of this concept in day to day life ? |
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Answer» (i) Torque plays the same role in rotational motion as force plays in linear motion. (ii) The direction of torque, `overset rarr(tau)` is given by right handed screw rule. `overset rarr(tau) _|_ overset rarr(r ) and overset rarr(tau) _|_overset rarr(F)`. (iii) From `oversetrarr(tau) = overset rarr(r ) xx overset rarr(F) = rF sin theta`, we find that when `theta = 0^(@) = zero`. In day to day life, it implies that when you align yourself along the force (i.e., your boss), the turning effect on you becomes zero, ity, say God, i.e., when you merge your wishes with the wishes of God, you are stable. |
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| 163. |
Fill in the blanks : (i) `overset rarr(tau) = overset rarr(r ) xx …` (ii) `overset rarr(L) = overset rarr(r ) xx …` |
| Answer» (i) `overset rarr(tau) = overset rarr(r ) xx overset rarr(F)` (ii) `overset rarr(L) = overset rarr(r ) xx overset rarr(p)` | |
| 164. |
Using expressisions for power and kinetic energy of rotational motion, derive the relation `tau = I alpha`, where letters have their usual meaning. |
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Answer» We know that power in rotational motion, `P = tau omega` ..(i) and `K.E.` of rotation, `E = (1)/(2)I omega^(2)` …(ii) As power=time rate of doing work in rotational motion, and work is stored in the body in the from of `K.E.` `:. P = (d)/(dt) (K.E.` of rotation) `= (d)/(dt) ((1)/(2)I omega^(2)) = (1)/(2)I xx 2 omega((d omega)/(dt))` Using (i), `P = tau omega alpha` or `tau = I alpha`, which is the required relation. |
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| 165. |
The correct relation between torwue `tau` and angular momentum `L` isA. `tau = (dL)/(dt)`B. `L = (d tau)/(dt)`C. `tau = L xx t`D. `L = tau xx t` |
| Answer» Correct Answer - A | |
| 166. |
A circular disc `X` of radius `R` is made from an iron plate of thickness `t`, and another plate `Y` of radius `4R` is made from an iron plate of thickness `t//4`. The ratio between moment of inertia `I_(Y)//I_(X)` isA. `32`B. `16`C. `1`D. `64` |
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Answer» Correct Answer - D `I_(X) = (1)/(2) ("mass") ("radius")^(2)` `= (1)/(2) [(piR^(2))t.rho] R^(2)` ltbr. `I_(Y) = (1)/(2) [(pi 10 R^(2)) (t.rho)/(4)](4R)^(2) = 64 I_(X)` |
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| 167. |
A disc of metal is melted and recast in the from of a soild sphere. What will happen to the moment of inertia anout a vertical axis passing through the centre ? |
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Answer» Moment of inertia will decrease, because `I_(d) = (1)/(2)m r^(2) and I_(s) = (2)/(5)m r^(2)`, the radius of sphere formed on recasting the disc will also decrease. |
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| 168. |
A soild sphere is made to roll down from the same height on two inclined planes having different angles of inclination. In which case will it take less time to reach the bottom ? |
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Answer» As two inclined planes have different inclinations, but same height, therefore, velocity on reaching the bottom along two inclines is the same. From `upsilon = u + at = 0 + at` Now, `upsilon` is same and `a sin theta :. t prop (1)/(sin theta)` i.e. time taken will be smaller when the spheres rolls the plane of higher inclination. |
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| 169. |
A circular disc of radius `r` is rolling without slipping on a horizontal surface. What is the ratio of the translational `KE` and rotational `KE` of disc ? |
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Answer» Correct Answer - `(2)` `KE` of translation, `KE_(T) = (1)/(2)mv^(2)` `KE` of rotation, `KE_(R ) = (1)/(2)I omega^(2)` `= (1)/(2) ((1)/(2)mr^(2)) omega^(2) = (1)/(4)mv^(2)` `(KE_(T))/(KE_(R )) = ((1)/(2)mv^(2))/((1)/(4)mv^(2)) = 2` |
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| 170. |
A spherical ball rolls on a table without slipping. Determine the percentage of its `K.E.` which is rotational. Moment of inertia of sphere `= (2//5) xx mass xx ("radius")^(2)`. |
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Answer» Total energy `= K.E.` of translation `+ K.E.` of rotation `(1)/(2)mv^(2) + (1)/(2)I omega^(2)` `E = (1)/(2)m v^(2) + (1)/(2) ((2)/(5) m r^(2)) omega^(2)` `= (1)/(2)m v^(2) + (1)/(5)m v^(2) = (7)/(10)m v^(2)` `(K.E. "of rotation")/("total energy") = (((1)/(2))I omega^(2))/(((7)/(10))m v^(2))` `=(((1)/(5))m v^(2))/(((7)/(10))m v^(2)) =(2)/(7) = (2)/(7) xx 100% = 28.57 %` |
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| 171. |
A cylinder of radius `R` and mass `M` rolls without slipping down a plane inclined at an angle `theta`. Coeff. of friction between the cylinder and the plane is `mu`. For what maximum inclination `theta`, the cylinder rolls without slipping ?A. `tan^(-1) mu`B. `tan^(-1) (3mu)`C. `tan^(-1) 2mu`D. `tan^(-1) .(3)/(2) mu` |
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Answer» Correct Answer - B As is know from theory, `mu = (tan theta)/(1 + mR^(2)//I) = (tan theta)/(1+2) = (tan theta)/(3)` `tan theta_(max) = 3mu` `theta_(max) = tan^(-1) (3mu)` |
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| 172. |
A solid cylinder at rest at top of an inclined plane of height `2.7 m` rolls down without slipping. If the same cylinder has to slide down a frictionaly inclined plane and acquires the same velocity as that acquired by centre of mass of rolling cylinder at the bottom of the incline, what should be the height of inclined plane ? |
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Answer» Here, `h_(1) = 2.7m, h_(2) = ?` When the solid cylinder rolls without slipping, its total kinetic energy `K = K_(t) + k_(r )` `= (1)/(2)m upsilon^(2) + (1)/(2)I omega^(2)` `= (1)/(2)m upsilon^(2) + (1)/(2)((1)/(2)m r^(2)) omega^(2)` `K = (1)/(4) m upsilon^(2) + (1)/(4)m upsilon^(2) = (3)/(4)m upsilon^(2)` But `K = mg h_(1) = (3)/(4)m upsilon^(2)` ..(i) In sliding down a height `h_(2)`. `mg h_(2) = (1)/(2)m upsilon^(2)` ...(ii) Dividing (ii) by (i), we get `(h_(2))/(h_(1)) = (1//2)/(3//4) = (2)/(3)` `h_(2) = (2)/(3)h_(1) = (2)/(3) xx 2.7 m = 1.8 m` |
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| 173. |
A `70kg` man stands in contact against the inner wall of a hollow cylindrical drum of radius `3m` rotating about its verticle axis. The coefficient of friction between the wall and his clothing is `0.15` . What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed? |
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Answer» Here, `m = 70 kg, r = 3` `n = 200"rpm", mu = 0.15, omega_("min") = ?` The necessary centripetal force `(m upsilon^(2)//r)` is provided by horizontal normal reaction `R` of the wall on the man, i.e.., `R = (m upsilon^(2))/(r ) = m r omega^(2)` The frictional force `f` acting upwards, balances the weight `(mg)` of the man. The man will remain struck to the wall after the floor is removed, provided. `f = mu R,i.e., mg le mu (m r omega^(2))` or `omega^(2) ge (g)/(mu r)` `:. omega_(min) = sqrt((g)/(mu r)) = sqrt((9.8)/(0.15 xx 3))` `= sqrt(21.77) = 4.67 rad//s` |
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| 174. |
A thin hollow cylinder, open at both ends and weighing `5kg` (a) slides with a speed of `10 m//s` without rotating (b) rolls with the same speed without slipping. Compare the kinetic energies of the cylinder in the two cases. |
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Answer» Here, `m = 5 kg, v = 10 m//s` (a) When cylinder slides without rotating, its energy is wholly translatory. `E_(1) = (1)/(2) m upsilon^(2) = (1)/(2) xx 5 xx 10^(2) = 250 J` (b) When cylinder rolls without slipping, it has both translational and rotational `K.E`. `E_(2) = (1)/(2) m upsilon^(2) + (1)/(2) I omega^(2)` But `I = m r^(2) and r omega = upsilon` `:. E_(2) = (1)/(2)m upsilon^(2) + (1)/(2)(m r^(2)) omega^(2) = (1)/(2) m upsilon^(2) + (1)/(2) m upsilon^(2) = m upsilon^(2) = 5 xx 10^(2) = 500 J :. (E_(1))/(E_(2)) = (250)/(500) = (1)/(2)` |
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| 175. |
Calculate the angular velcity of the minute`s hand of a clock.A. `pi//30`B. `pi//1800`C. `2pi//30`D. `(2pi)/(1800)` |
| Answer» Correct Answer - B | |
| 176. |
If the earth expands suddenly to twice its diamteter, what would be the length of the day ? |
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Answer» Here, `D_(2) = 2D_(1) :. R_(2) = 2 R_(1)`. `T_(1) = 1 day = 24 hr, T_(2) = ?` According to the law of conservation of angular momentum, `I_(2)omega_(2) = I_(1)omega_(1)` `(2)/(5)mR_(2)^(2) ((2pi)/(T_(2))) = (2)/(5)mR_(1)^(2) ((2pi)/(T_(1)))` `T_(2) = T_(1) ((R_(2))/(R_(1)))^(2) = 24 (2)^(2) = 96 hours` |
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| 177. |
If the earth suddenly contracts to one third of its present size, calculate by how much would the day be shortened ? |
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Answer» Here, `R_(2) = (1)/(3)R_(1), T_(1) = 24 hrs, T_(2) = ?` According to the principle of conservation of angular momentum `I_(2)omega_(2) = I_(1)omega_(1)` `I_(2) ((2pi)/(T_(2))) = I_(1) ((2pi)/(T_(1)))` `T_(2) = (I_(2))/(I_(1)) xx T_(1) = ((2)/(5)MR_(2)^(2))/((2)/(5)MR_(1)^(2)) xx T_(1)` `T_(2) = T_(1) xx ((R_(2))/(R_(1)))^(2)` `= 24 xx ((1)/(3))^(2) = 2 hr 40 min`. `:. T_(1) - T_(2) = 24 hr - 2hr 40 min` `= 21 hr 20 min` |
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| 178. |
What would be the length of the day, if earth were to shrink suddenly to `1//64th` of its original volume ?A. `24 h`B. `12 h`C. `6 h`D. `1.5 h` |
| Answer» Correct Answer - D | |
| 179. |
The rotational analgue of force isA. momentumB. angular momentumC. torqueD. none of these |
| Answer» Correct Answer - C | |
| 180. |
Choose the correct alternatives :A. For a general rotational motion, angular momentum `L` and angular velocity `omega` need not be parallel.B. For a rotational motion about a fixed axis, angular momentum `L` and angualr velocity `omega` are always parallel.C. For a general translational motion, mometum `p` and velocity `v` are always parallel.D. For a general translational motion, acceleration a and velocity v are always parallel. |
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Answer» Correct Answer - A::C From the study of theory, we know that for general rotational motion, angualr momentum `overset rarr(L)` and angular velocity `overset rarr(omega)` need not be parallel. Again, for a general translation motion, linear momentum `overset rarr(p)` and linear velocity `overset rarr(v)` are always parallel. This is because `overset rarr(p)` is directed along `overset rarr(v)` only. |
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| 181. |
Torques of equal magnitude are applied to hollow cylinder and a solid sphere, both having the same mass and same radius. The cylinder is free to rotate aabout its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. which of the two will aquire a greater angular speed after a given time ? |
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Answer» If `M` is mass and `R` is radius of the hollow cylinder and the solid sphere, then `M.I.` of hollow cylinder about its axis of symmetry, `I_(1) = MR^(2)`, and `M.I.` of solid sphere about an axis through its centre, `I_(2) = (2)/(5)MR^(2)` Torque applied, `tau = I_(1) alpha_(1) = I_(2) alpha_(2) :. (alpha_(2))/(alpha_(1)) = (I_(1))/(I_(2)) = (MR^(2))/((2)/(5) MR^(2)) = (5)/(2) :. alpha_(2) gt alpha_(a)` From `omega = omega_(0) + alpha t`, we find that for given `omega_(0) and t, omega_(2) gt omega_(1)` i.e. angular speed of solid sphere will be greater than the angular speed of hollow cylinder. |
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| 182. |
Figure shows two identical particles `1` and `2`, each of mass `m`, moving in opposite directions with same speed `vec V` along parallel lines. At a particular instant, `vec r_1` and `vec r_2` are their respective position vectors drawn from point `A` which is in the plane of the parallel lines. Which of the following is the correct statement ? .A. Angular momentum `overset rarr(L_(1))` of particle 1 about `A` is `overset rarr(L_(1)) = mv (d_(1)) o.`B. Angular momentum `overset rarr(L_(2))` of particle 2 about A is `overset rarr(L_(1)) = mv r_(2) o.`C. Total angular momentum of the system about `A` is `overset rarr(L) = mv (r_(1) + r_(2))o.`D. Total angular momentum of the system about `A` is `overset rarr(L) = mv (d_(2) - d_(1)) ox` `o.` respresents a unit vector coming out of the page. ox represents a unit vector going into the page. |
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Answer» Correct Answer - A::D As angular momentum = linear momentum multiplied by perpendicular distance of particle from the axis of rotation, therefore, for particle 1, `overset rarr(L_(1)) = mv (d_(1)) o.`, perpendicular to plane of page and directed upwards, and for particle 2. `overset rarr(L_(2)) = mv (d_(2)) ox`, perpendicular to plane of page and directed inwards. `:.` Total angular momentum of the system, `overset rarr(L) = overset rarr(L_(2)) - overset rarr(L_(1)) = mv(d_(2) - d_(1)) ox` It would be perpendicualr to plane of page and directed inwards. |
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| 183. |
A solid sphere of mass `3 kg` and radius `2m` is free to rotate about an axis passing through its centre. Find a constant tangential force `F` required to rotate the sphere with a velocity of `10 rad//s` in `2 sec`. Starting from rest. Also, find the number of rotations made by the sphere in that time interval. |
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Answer» Here, `m = 3kg, r = 2m omega_(0) = 0` `omega = 10 rad//s, t = 2s` `alpha = (omega - omega_(0))/(t) = (10 - 0)/(2) = 5 rad//s^(2)` `theta = omega_(0)t + (1)/(2)alpha t^(2) = 0 + (1)/(2)xx5 xx 2^(2) = 10 rad` Now, `tau = F xx r = I alpha - (2)/(5)m r^(2) alpha` `F = (2)/(5)m r alpha = (2)/(5) xx 3 xx2 xx 5 = 12 N` Number of revolutions `= (theta)/(2pi) = (10)/(2pi) = (5)/(pi)` |
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| 184. |
The centre of gravity of a loaded texi is `1.5 m` above the ground, and the distance between the wheels is `2m`. What is the maximum speed with which it can go round an unbanked curve of radius `100 m` without being turned upside down. What minimum value of coefficient of friction is needed at this speed ? |
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Answer» Applying condition for rotational equilibrium, `(mv^(2))/(r ) xx h = mg (x), where 2x = 2m, h = 1.5 m` `v = sqrt((g r x)/(h)) = sqrt((9.8 xx 100 xx 1)/(1.5)) = 25.56 m//s` `mu = v^(2)//r g = ((25.56)^(2))/(100 xx 9.8) = 0.67`. |
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| 185. |
The vector product of two vectors `overset rarr(A) and overset rarr(B)` is another………..whose magnitude is equal to…………..and………….between them. |
| Answer» vector `overset rarr(c)` , vector of magnitude of `overset rarr(A) and overset rarr(B)`, sine of smaller angle `theta`. | |
| 186. |
A disc of moment of inertia `I_(1)` is rotating freely with angular speed `omega_(1)` when another non-rotating disc of moment of inertia `I_(2)` is dropped on it. The two discs then rotate as one unit. Find the final angular speed. |
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Answer» Here, initial angular momentum `L_(1) = I_(1) omega_(1)` Let `omega` be the final angular velocity of the combination. Therefore, final angular momentum. `L_(2) = (I_(1) + I_(2)) omega`. As `L_(2) = L_(1)` `:. (I_(1) + I_(2)) omega = I_(1) omega_(1)` `omega= (I_(1) omega_(1))/(I_(1) + I_(2))` |
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| 187. |
A circular disc of moment of inertia `I_(t)` is rotating in a horizontal plane about its symmetry axis with a constant angular velocity `omega_(i)`. Another disc of moment of inertia `I_(b)` is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed `omega_(f)`. Calculate the energy lost by the initially rotating disc due to friction.A. `(1)/(2)(I_(t)I_(b))/((I_(t) + I_(b))) (i)_(i)^(2)`B. `(1)/(2) (I_(t)^(2))/((I_(t) + I_(b))) (i)_(i)^(2)`C. `(I_(b)-I_(t))/((I_(t) + I_(b))) (i)_(i)^(2)`D. `(1)/(2) (I_(b)-I_(t))/((I_(t) + I_(b))) (i)_(i)^(2)` |
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Answer» Correct Answer - A As no external torque is acting, therefore, `I_(t)(i)_(i) = (I_(t) + I_(b)) (i)_(f)` `(i)_(f) = (I_(t)(i)_(i))/(I_(t) + I_(b))` Energy lost `= DeltaK = E_(1) - E_(2)` `DeltaK = (1)/(2)I_(t) (i)_(i)^(2) - (1)/(2)(I_(t) + I_(b))(i)_(f)^(2)` `= (1)/(2)I_(t) (i)_(i)^(2) - (1)/(2) ((I_(t) + I_(b))I_(t)^(2)(i)_(i)^(2))/((I_(t) + I_(b)^(2))` `= (1)/(2)I_(t)(i)_(t)^(2) [1 - (I_(t))/(I_(t) + I_(b))]` `= (1)/(2)I_(t)(i)_(i)^(2) (I_(b))/((I_(t) + I_(b)))` |
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| 188. |
A circular disc of moment of inertia `I_(t)` is rotating in a horizontal plane about its symmetry axis with a constant angular velocity `omega_(i)`. Another disc of moment of inertia `I_(b)` is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed `omega_(f)`. Calculate the energy lost by the initially rotating disc due to friction. |
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Answer» Here, `I_(i) = I_(t)` Let `omega_(i)` be the initial angular speed `I_(f) = (I_(t) + I_(b))` Let `omega_(f)` be the final angular speed of the system. According to the law of conservation of angular momentum, `I_(f)omega_(f) = I_(i) omega_(i)` `omega_(f) = (I_(t)omega_(i))/(I_(t) + I_(b))` ..(i) Energy lost `= K_(f) - K_(i) = (1)/(2)I_(f) omega_(f)^(2) - (1)/(2)I_(i) omega_(i)^(2)` `= (1)/(2)(I_(t) + I_(b)) omega_(f)^(2) - (1)/(2)I_(t)omega_(t)^(2)` using (i),we get Energy lost `= (1)/(2) (I_(t) + I_(b)) ((I_(t)omega_(i))/(I_(t) + I_(b)))^(2) - (1)/(2)I_(t) omega_(i)^(2)` `= (1)/(2)I_(t) omega_(i)^(2) [(I_(t))/(I_(t) + I_(b)) - 1] = - (1)/(2) (I_(b)I_(t)omega_(i)^(2))/((I_(t) + I_(b)))` Energy lost `=- (1)/(2)I_(b) omega_(i) omega_(f)` [using (i)] |
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| 189. |
A child is sitting at one end of a long trolley moving with a uniform speed `v` on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed `u` (w.r.t. trolley), the speed of the centre of mass of the system will. |
| Answer» The speed of the centre of mass of the system (trolley + child) shall remain unchanged, when the child gets up and runs about on the trolley in any manner. This is because forces involed in the exercise are purely internal i.e. from within the system. No external force acts on the system and hence there is no change in velocity of mass of the system. | |
| 190. |
Three point masses of `1 kg, 2kg and 3 kg` lie at `(1,2), (0,-1) and (2, -3)` respectively. Calculate the co-ordinates of the centre of mass of the system. |
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Answer» Here `m_(1) = 1 kg, m_(2) = 2 kg, m_(3) = 3 kg` `x_(1) = 1, y_(1) = 2 , x_(2) = 0, y_(2) =- 1`, `x_(3) = 2, y_(3) =- 3` Co-ordinates of centre of mass `(x,y)` are given by `x = (m_(1)x_(1) + m_(2)x_(2) + m_(3)x_(3))/(m_(1) + m_(2) + m_(3))` `= (1(1) + 2(0) + 3(2))/(1 + 2+ 3) = (7)/(6)` `y = (m_(1)y_(1) + m_(2)y_(2) + m_(3)x_(3))/(m_(1) + m_(2) + m_(3))` `= (1(2) + 2(-1) + 3(-3))/(1 + 2+ 3) = (-3)/(2)` |
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| 191. |
Does centre of mass of a rigid body lie always on the body ? |
| Answer» No, it is not necessary. The centre of mass of a rigid body may or may not lie on the body. | |
| 192. |
Statement-1 : Torque is a vector whose direction is along the applied froce. Statement-2 : `overset rarr(tau) = overset rarr(r ) xx overset rarr(F)`A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true but Statement-2 is not correct explanation of Statement-1.C. Statement-1 is true, but statement-2 is false.D. Statement-1 is false, but statement-2 is true. |
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Answer» Correct Answer - D Torque is a vector quantity, whose direction is perpendicualr to the direction of `overset rarr(r )` and `overset rarr(F)`. This is as per right handed screw rule. Statement-1 is false but statement-2 is true. |
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| 193. |
Statement-1 : The centre of mass of a body may lie where there is no mass. Statement-2 : The centre of mass has nothing to do with the mass.A. Statement-1 is true, Statement-2 is true , Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true but Statement-2 is not correct explanation of Statement-1.C. Statement-1 is true, but statement-2 is false.D. Statement-1 is false, but statement-2 is true. |
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Answer» Correct Answer - C Statement-1 is true, but the statement-2 is not true. Inface, the centre of mass is related to the distribution of mass of the body. Option A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion Both the Assertion and reason statements are correct and reason explains the assertion very well. For example : Center of mass of the ring lies at its geometric centre where even there is no mass of any part of ring.
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| 194. |
Two satellites of equal masses, which can be considered as particles are orbiting the earth at different heights. Will their moment of inertia be same or different ? |
| Answer» Moments of inertia of the two satellites will be different. This is because, `I = mass xx (disatnce)^(2)`. For the satellite revolving at a greater height, distance from the axis of rotation is larger. Therefore, its moment of inertia is larger. | |
| 195. |
Determine the angular momentum of a car of mass `200 kg` moving in a circular track of diameter `100 m` with a speed of `40 m//s`. |
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Answer» Here, `L = ?, m = 2000 kg , r =(100)/(2)m = 50 m`, `v = 40 m//s, L = mv r = 2000 xx 40 xx 50` `= 4 xx 10^(6) kg m^(2) s^(-1)` |
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| 196. |
The moment of linear momentum is calledA. torqueB. coupleC. angular momentumD. none of the above |
| Answer» Correct Answer - C | |
| 197. |
A particle moves on a straight line with a uniform velocity. The angular momentum of the particles isA. Always zeroB. zero about a point on the stright lineC. zero about a point away from the straight lineD. constant always about a given point not on the line |
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Answer» Correct Answer - B::D As `L = mv r`, therefore `L = 0`, when `r = 0` i.e. when point lies on the straight line. When a given point does not lie on the line, `r =` consatnt. Therefore, `L = mv r =` consatnt. |
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| 198. |
Find the scalar and vector paroducts of two vectors. `a=(3hat(i)-4hat(j)+5hat(k)) " and " b=(-2hat (i) +hat(j)-3hat(k))` |
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Answer» `a.b=(3hat(i)-4hat(j)+5hat(k)).(-2hat(i)+hat(j)-3hat(k))` `=-6-4-15` `=-25` `axxb=|{:(" "hat(i)," "hat(j)," "hat(k)),(" "3 ,-4," "5),(-2," "1,-3):}|=7hat(i)-hat(j)-5hat(k)` `bxxa=-7hat(i)+hat(j)+5hat(k)` |
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| 199. |
Can a body in translatory motion have angular momentum ? Explain. |
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Answer» Yes, a body in translatory motion shell have angular momentum, unless the fixed point about which angular momentum is taken lies on the line of motion of the body. This follows from `|L| = r p sin phi L = 0`, only when `phi = 0^(@) or phi = 180^(@)`. |
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| 200. |
Angular momentum of a particle about a given axis is …………….of ………….and………….of position vector of the particle. |
| Answer» twice the product , mass of particle , areal velocity. | |