596 + Interview Questions in LAWS-OF-MOTION IN GENERAL AWARENESS Page 4 InterviewSolution

151.	A light string passing over a smooth light pulley connects two blocks of masses `m_1` and `m_2` (vertically). If the acceleration of the system is `g//8`, then the ratio of the masses isA. `8:1`B. `9:7`C. `4:3`D. `5:3`
Answer» Correct Answer - b In the given system, `a =(m_(1) -m_(2)g)/(m_(1) + m_(2)) =(g)/(8)` `:. (m_(1)-m_(2))/(m_(1) +m_(2)) =(1)/(8)` `8m_(1) -8m_(2) = m_(1) + m_(2)` `7 m_(1) = 9m_(2)` `(m_(1))/(m_(2)) = (9)/(7)` .

Discussion

152.	Two masses `m_1=5kg` and `m_2=4.8kg` tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when left free to move? A. (a) `5m//s^2`B. (b) `9.8m//s^2`C. (c) `0.2m//s^2`D. (d) `4.8m//s^2`
Answer» Correct Answer - C Acceleration `a=((m_1-m_2)/(m_1+m_2))g` `=((5-4.8)xx9.8)/(5+4.8)m//s^2=0.2m//s^2`

Discussion

153.	Ten one rupee coins are put on stop of one another on a table Each coin has a mass m kg Give the magnitude and direction of (a) the force on the 7 th coin (counted from the botton ) due to all coins above it (b) the force on the 7 th coin by the eighth coin and (c) the reaction of the sixth coin on the seventh coin.
Answer» (a) The force on 7 th coin is due to weight of the three coins lying above it Therefore ` F = (3 m ) kg f = (3 mg ) N ` where g is acceleration due to gravity . This force acts vertically downwards . (b) The eighth coin is alredy under the weight of two coins above it and it has its own weight too . Hence force on 7 th coin due to 8 th coin is sum of the two forces i . e ` F = 2 m + m = (3 m ) kg f = (3 mg ) N ` The force acts vertically downwards . (c) The sixth coin is under the weight of four coins above it Reaction ` R = - F = - 4 m (kg f ) = - (4 mg ) N ` Minus sigh indicates that the reaction acts vertically upwards opposite to the weight .

Discussion

154.	Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m . Give the magnitude and direction of (a) the force on the `7^(th)` coin (counted from the bottom due to all the coins on its top. (b) the force on the `7^(th)` coin by the eight coin. (c) the reaction of the `6^(th)` coin one the `7^(th)` coin.
Answer» (a) The force on 7th coin is due to weight of the three coins lying above it. Therefore, F = (3m)kgf = (3mg) N where g is acceleration due to gravity . This force acts vertically downwards . (b) The eighth coin is already under the weight of two coins above it and it has its own height too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e. F = 2m + m = (3m) kg f = (3mg)N. The force acts vertically downwards . (c ) The sixth coin is under the weight of four coins above it . Reaction , R = - F = - 4m(kg) = -(4mgf)N Minus sign indicates that the reaction acts vertically upwards , opposite to the weight .

Discussion

155.	A ball of mass `20` gram hits a smooth wall at an angle of `45^@)` with a velocity of `15m//s` If ball rebounds at `90^(@)` to the direction of incidence, calculate the impulse received by the ball.
Answer» Correct Answer - ` 0.3 sqrt2 kg ms^(-1)` .

Discussion

156.	A ball weighing `10g` hits a hard surface vertically with a speed of `5m//s` and rebounds with the same speed The ball remains in contact with the surface speed The ball remains in contact with the surface for `0.01s` The average force exerted by the surface on the ball is .A. `100N`B. `10N`C. `1N`D. `0.1N`
Answer» Correct Answer - b Here, `m = 10 g = (10)/(1000)kg = 0.01kg` `upsilon_(1) = -5m//s, upsilon_(2) = + 5m//s, t = 0.01sec` According to impulse momentum theorem Impulse = change in momentum `F _(av) xx t = m upsilon_(2) - m (-upsilon_(1)) =m upsilon_(2) + m upsilon_(1)` `F_(av) xx 0.01 = 0.01 xx 5 + 0.01 xx 5` `F_(av) = (0.1)/(0.01) = 10N` .

Discussion

157.	A monkey of mass `20kg` is holding a vertical rope. The rope will not break when a mass of `25kg` is suspended from it but will break it the mass exeeds `25kg` . What is the maximum acceleration with which the monkey can climb up along the rope? `(g=10m//s^(2))` .A. `25m//s^(2)`B. `2.5m//s^(2)`C. `5m//s^(2)`D. `10m//s^(2)`
Answer» Correct Answer - B

Discussion

158.	A body takes time `t` to reach the bottom oa a smooth inclined plane of angle `theta` with the horizontal. If the plane is made rought, time taken now is `2t`.The coefficient of friction of the rough surface isA. `(3)/(4) tan theta`B. `(2)/(3) tan theta`C. `(1)/(4) tan theta`D. `(1)/(2) tan theta`
Answer» Correct Answer - A `S = (1)/(2) at^(2)` `:. t= sqrt((2S)/(a)` or `t prop (1)/(sqrt(a))` `:. (t_(1))/(t_(2)) = ((a_(2))/(a_(2)))` or `(t)/(2t) = ((g sin theta - mug cos theta)/(g sin theta ))` On solving this equation , we get `mu (3)/(4) tan theta`

Discussion

159.	The force on a rocket moving with a veloctiy 300 m/s is 210N. The rate of consumption of fuel of rocket isA. `0.07 kg s^(-1)`B. `1.4kg s^(-1)`C. `0.7kg s^(-1)`D. `10.7kg s^(-1)`
Answer» Correct Answer - C Force `=(d)/(dt)` (momentum) `(d)/(dt)("mv")=v((dm)/(dt)) Rightarrow 210=300((dm)/(dt))` rate of combustion, `(dm)/(dt)=(210)/(300)=0.7 kg s^(-1)`

Discussion

160.	A rocket motor consumes 100 kg of fuel per second exhausting it with a speed of `6 xx 10^(3) ms^(-1)` What thrust is exerted on the rocket ? What will be the velocity of the rocket at the instant its mass is reduced to `(1 // 40)` of its initial mass ? Take initial velocity of rocket as zero . Neglect gravity .
Answer» Correct Answer - (i) `6xx10^(5)N` (ii) `22.13xx10^(3)ms^(-1)`

Discussion

161.	What will be the maximum speed of a car on a road turn of radius 30 m, if the coefficient of friction between the tyres and the road is 0.4? (Take `g=9.8m//s^(2)`)A. 10.84 m/sB. 9.84 m/sC. 8.84 m/sD. 6.84 m/s
Answer» Correct Answer - A

Discussion

162.	A shell is fired from a cannon, it explodes in mid air, its totalA. momentum increasesB. momentum decreasesC. KE increasesD. KE decreases
Answer» Correct Answer - B

Discussion

163.	If the force on a rocket moving moving with a velocity of 300 m/s is 345 N, then the rate of combustion of the fuel isA. 0.55 kg/sB. 0.75 kg/sC. 1.15 kg/sD. 2.25 kg/s
Answer» Correct Answer - C

Discussion

164.	A rocket has a mass of `2 xx10 ^(4)` kg of which half is fuel Assume that the fuel is consumed at a constant rate as the rocket is fired and there is contact thrust of `5 xx 10^(6)` N neglecting air resistance and any possible variation of g comput (i) the intial acceleration (ii) acceleration just when the whole fuel is consumed .
Answer» Here, mass of rocket `m = 2xx 10^(4) kg` mass of fuel `m = (1)/(2) (2xx10^(4)) kg = 10^(4) kg` upthrust `= 5 xx 10^(6) N` (i) Net force on the rocket F = up thrust - weight (mg) `5 xx 10^(6) -2xx 10^(4) xx 10` ` = 10^(5) (50 -2) = 48 xx 10^(5) ` `:.` Initial acceleration `= (F)/(m) = (48xx 10^(5))/(2xx10^(4)) = 240 m//s^(2) ` When the fuel has been completly exhausted , remaining mass of rocket F = upthrust - weight (m' g) `5 xx 10^(6) - 10^(4) xx 10 = 49 xx 10^(5) ` `:.` Acceleration = `(F)/(m) = (49 xx 10^(5))/(10^(4)) = 490 m//s^(2)` .

Discussion

165.	A car moving on a level rosd. A pendulum suspended from the ceiling makes an angle of `10^(@)` with the vertical. Find the acceleration of the car. Take `g=10ms^(-2)`.
Answer» Correct Answer - `1.763ms^(-2)`

Discussion

166.	A body of weight `200N` is suspended with the help of strings as show in Find the tensions in the strings .
Answer» Correct Answer - `T_(1)=146.4N, T_(2)=179.3N`

Discussion

167.	A mass of 10 kg is suspended vertically by a rope of length 2 m from a ceiling. A force of 60 N is applied at the middle point of the rope in the horizontal direction, as shown in Fig. Calculate the angle the rope makes with the vertical. Neglect the mass of the rope and take `g =10ms^(-2)`.
Answer» Correct Answer - `31^(@)`

Discussion

168.	A mass of 10 kg is suspended from a string, the other end of which is held in hand Find the tension in string when the hand is moved up with a uniform acceleration of `2ms^(-1)`. Given `g = 10 ms^(-1)`.
Answer» Correct Answer - 120 N

Discussion

169.	A 30 kg Sell is flying at `36 m//s`. When the shell explodes into two parts of `12 kg` and `18 kg` , the lighter part stops, and heavier part files on . What is the velocity of heavier part ? .
Answer» Here, `M = 30 kg , u = 36 m//s` `m_(1) = 12 kg , upsilon_(1) = 0 , m_(2) = M- m_(1) = 30 - 12 = 18 kg , upsilon_(2) = ?` From the law of conservation of lincear momentum , `m_(1) upsilon + m_(2) upsilon_(2) = M xx u , 0+ 18 upsilon_(2) = 30 xx 36 upsilon_(2) = (30 xx 36 )/(18)=60 m//s` .

Discussion

170.	A weightless thread can bear tension upto `3.7 kg` wt A stone of mass `500g` is tied to it and revolves in a verticle circle of radius `4m` What will be the maximum angular velocity of the stone if `g = 10 m//s^(2)` .
Answer» Correct Answer - `4rad//s` . Here, `T_(max) = 3.7kg wt = 37 N` `m = 500 g = 0.5kg` `r = 4 m, omega = ? g = 10m//s^(2)` As `T _(max) = (m upsilon^(2))/(r)+ mg` `:. (m upsilon^(2))/(r) = T_(max) - mg = 37 - 0.5 xx 10 =32` `upsilon^(2) =(32r)/(m) =(32 xx 4)/(0.5) = 256` `upsilon = sqrt(256) = 16m//s` ` omega = (upsilon)/(r) = (16)/(4) = 4rad//s` .

Discussion

171.	The strings of a parachute can bear a maximum tension of `72 kg` wt . By what minimum acceleration can a person of 90 kg descend by means of this parachute ?
Answer» Here ` T = 72 kg wt . = 72 xx 9.8 N ` `m = 90 kg , a = ? ` for the person to descend , `T = (g-a)` `72 xx 9.8 = (9.8-a ) (72xx9.8)/(90)=9.8-a 0.8 xx 9.8 = 9.8 - a` `a = 9.8 - 0.8 xx 9.8 = 9.8 (1-0.8)= 1.96 m//s^(2) ` .

Discussion

172.	A body subjected to three concurrent force is found to be in equilibrium. The resultant of any two forceA. is equal to third forceB. is equal to third forceC. is collinear fifth the third forceD. all of these
Answer» Correct Answer - D For equilibrium under the effect of the three concurrent force all the properties mentioned are required.

Discussion

173.	A small block of mass 1kg is placed over a plank of mass 2kg. The length of the plank is 2 m. coefficient of friction between the block and the plnak is 0.5 and the ground over which plank in placed is smooth. A constant force F=30N is applied on the plank in horizontal direction. the time after which the block will separte from the plank is `(g=10m//s^(2))` A. 0.73sB. 1.2sC. 0.62sD. 1.6s
Answer» Correct Answer - a

Discussion

174.	Two block `A` and `B` of masses `1kg` and`2kg` respectively are connected by a string, passing over a light frictionless pulley `B` as shown. Another string connect the center of pulley. Both the blocks are resting on a horizontal floor and the pulley is help such that string remains just taut. At moment `t = 0`, a force `F = 20 t` starts acting on the pully along vertically upwards direction as shown in figure.Calculate (a) velicity of `A` when `B` loses contact with the floor. (b) height raised by the pulley upto that instant. (Take =`g =10 m//s^(2))`
Answer» (a) Let `T` be the tansion in the string. Then, `2T =20 t` `T =20 t` newton Let the block `A` loses its contact with the floor at time `t =t_(1)`. This happens when the tension in string becomes equal to the weight of `A` . Thus, `T =mg` or `10t_(2) =1xx10` or `t_(2) =1s` ...(i) Similarly, for block `B` we have `10t_(2) =2xx10` or `t_(2) =2s` ...(ii) i.e. the block `B` loses contact at `2s`. For block `A`, at time `t` such that `tget_(1)` let a be its acceleration in upward direction.Then, `10t-1xx10 =1xxa = (dv//dt)` `du =10(t-1)dt` ...(iii) Integrating this expression, we get `int_(0)^(v) dv = 10 int_(1)^(t) (t-1) dt` or `v = 5t^(2)-10t+5` ...(iv) Substituting `t = t_(2) = 2s` `v = 20 - 20 + 5 = 5m//s` ...(v) (b) From Eq. (iv). `dy = (5t_(2)-10t + 5)dt` ...(vi) when, `y` is the vertical displacement of block `A` at time `t (ge t_(1))`. Integrating, we have `int_(y = 0)^(y =h) =int_(t = 1)^(t =2) (5t^(2)-10t + 5)dt` `h = 5[(t^(3))/(3)]_(1)^(2) -10 [(t^(2))/(2)]_(1)^(2) + 5[t]_(1)^(2) = (5)/(3)m` :. Height raised by pulley upto that instant `=(h)/(2) = (5)/(6) m`

Discussion

175.	A block of mass 1 kg is stationary with respect to a converyor belt that is accelerationg with `1 m//s^(2)` upward at an angle of `30^(@)` as shown in figure . Which of the following statement (s) is /are correct ? `(g=10m//s^(2))` A. Force of friction on the block is 1 N upwardsB. Force of friction on the block is 1.52 N upwardsC. Contact force between the block and the belt is 10.5 ND. Contact force between the block and belt is `5sqrt(3) `N
Answer» Correct Answer - c

Discussion

176.	In the figure shown , all the string are massless and friction is absent everywhere. Choose the correct options. A. `T_(1) gt T_(3)`B. `T_(3) gt T_(1)`C. `T_(2) gt T_(1)`D. `T_(2) gt T_(3)`
Answer» Correct Answer - B::C::D Net pulling force = 0 `rArr a = 0` `T_(1) = 1 xx g = 10 N` `T_(3) = 2 xx g = 20 N` `T_(2) = 20 + T_(1) = 30 N`

Discussion

177.	A body of mass M at rest explodes into three pieces, two of which of mass M//4 each are thrown off in prependicular directions eith velocities of `3//s` and `4m//s` respectively. The third piece willl be thrown off with a velocity ofA. `1.5 ms^(-1)`B. `2 ms^(-1)`C. `2.5 ms^(-1)`D. `3 ms^(-1)`
Answer» Correct Answer - C ( c) Momentum of one piece` =(M)/(4)xx3` Momentum of other piece`=(M)/(4)xx4` `therfore` Resultant momentum`=sqrt((9M^(2)/(16)+M^(2))=(5M)/(4)` The third piece should also have the same momentum. Let its velocity be v, then `(5M)/(4)=(M)/(2)xxv impliesv=(5)/(2)=2.5 ms^(-1)`

Discussion

178.	A body of mass 1 kg at rest explodes into three fragments of masses in the ratio 1 : 1 : 3 . The two pieces of equal mass fly in mutually perpendicular directions with a speed of `30 m//s` each . What is the velocity of the heavier fragment ?
Answer» As total mass is 1 kg and fragments masses are in the ratio 1 : 1 : 3 , therefore , ` m_(1) = m_(2) = (1)/(5) kg and m_(3) = (3)/(5) kg ` also, ` upsilon_(1) = upsilon_(2) = 30 m//s , upsilon_(3) = ` ? According to the principal of conservation of linear momentum ` vec(p_(1))+ vec(p_(2))+ vec(p_(3))=0 ` ` vec(p_(3))+ vec(p_(1))+ vec(p_(2))=sqrt(p_(1)^(2)+p_(2)^(2))` ` m_(3) upsilon =m_(1) sqrt(upsilon_(2)^(1)+ upsilon_(2)^(2))=(1)/(5)sqrt(30^(2)+30^(2)` ` (3)/(5)upsilon_(3) =6 sqrt2 ` ` upsilon_(3) = (6 sqrt2xx5 )/(5)=10 sqrt2 = 14 . 14 m//s ` .

Discussion

179.	A bomb at rest explodes into three fragments of equal massses Two fragments fly off at right angles to each other with velocities of `9m//s` and `12//s` Calculate the speed of the third fragment .
Answer» Correct Answer - `-15m//s` . Let m be the mass of each fragment `:. P_(1) = mxx 9,p_(2) = m xx 12` As`vec(p_(1))` and` vec(p_(2))` are perpendicular to each other therefore resultant momentum of the two fragments `p = sqrt(p_(1)^(2) + p_(2)^(2)) = sqrt((9m)^(2) +(12m)^(2)) = 15m` According to the principle of conservation of linear momentum `vec(p) + vec(p_(3)) = 0` `vec(p_(3)) = - vec(p) = - 15 m ` `m xx upsilon_(3) = - 15m` `upsilon_(3) = -15m//s` Negative sign implies that third fragment will fly in a direction opposite to the direction of resultant momentum of the two fragments.

Discussion

180.	A bomb explodes in mid air into two equal fragments . What is the angle between their directions of motion ?
Answer» `180^(@)` as the two fragments would move in exactly opposite direction .

Discussion

181.	Impending state of mation is a critical border line between static and dynamic states of a body A block of mass m is supported on a rough vertical wall by applying a rough vertical wall by appying a force F as shwon in figure coefficent of static fricition between block and wall is `mu` The block under the influenace F sin`theta` may have a tendency to move upwart or it may be assumed that F `sin theta` justprevents downward fall of th block read the above prevents downward fall of the block .Read the above passage carefully and answer the following quesstions. The minimum value of force F required to keep the block stationary isA. `(mg)/(mucostheta)`B. `(mg)/(sin theta+mucostheta)`C. `(mg)/(sin theta-mucostheta)`D. `(mg)/(mu tantheta)`
Answer» Correct Answer - b

Discussion

182.	Impending state of mation is a critical border line between static and dynamic states of a body A block of mass m is supported on a rough vertical wall by applying a rough vertical wall by appying a force F as shwon in figure coefficent of static fricition between block and wall is `mu` The block under the influenace F sin`theta` may have a tendency to move upwart or it may be assumed that F `sin theta` justprevents downward fall of th block read the above prevents downward fall of the block .Read the above passage carefully and answer the following quesstions. the value of F for which friction force between the block and the wall is zero .A. mgB. `(mg)/(sintheta)`C. `(mg)/(costheta)`D. `(mg)/(tantheta)`
Answer» Correct Answer - b

Discussion

183.	A system consisting of a man on platfrom is in equilibrium. Masses of man and platiform are equal. The platform is in equilibrium due to two massless string is pulled by man resting on platform. Find the ratio of tension in left string to the tension in right string. A. `5//6`B. `4//3`C. `6//5`D. `3//4`
Answer» Correct Answer - c

Discussion

184.	A circular disc with a groove along its diameter is placed horizontally on a rough surface. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove and horizontal surface in contact is `mu=2/5`. The disc has an acceleration of `25m//s^2` towards left. Find the acceleration of the block with respect to disc. Given `cos theta=4/5`, `sin theta=3/5`.
Answer» Correct Answer - A::B Applying pseudo force `ma` and resolving it. Applying `F_(n et)=ma_r` `ma cos theta-(f_1+f_2)=ma_r` `macostheta-muN_1-muN_2=ma_r` `ma cos theta - mu ma sin theta - mu mg =ma_r` `implies a_r=acostheta-muasintheta-mug` `=25xx4/5-2/5xx25xx3/5-2/5xx10=10m//s^2`

Discussion

185.	These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following four responses Assertion A body of mass 10 kg is placed on a rough inclined surface `(mu=0.7)`. The surface is inclined to horizontal at angle `30^(@)`. Acceleration of the body down the plane will be zero. Reason Work done by friction is always negative.
Answer» Correct Answer - C ( c) `mg sin theta=10xx10xx(1)/(2)=50N` `mu mg cos theta=0.7xx10xx10xx(sqrt(3))/(2)=60.62N`

Discussion

186.	These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following four responses Assertion Earth is an inertial frame. Reason A frame in motion is sometimes an inertial frame and sometimes a non-inertial.
Answer» Correct Answer - D (d) Due to rotation of earth it is non-inertial. A frame moving with constant velocity is inertial.

Discussion

187.	A block of mass m is placed on an inclined plane with angle of inclination `theta` let N, `f_(L)` and F respectively represent the normal reaction limiting force of frcition and the net force down the inclined plane. Let `mu` be the coefficient of friction. the variations of N, `f_(L)` and F with `theta` is indicated by plotting graphs as shown its figure. Then, curves 1,2 and 3 respectively represent. A. N,F and `f_(L)`B. F, `f_(L)` and FC. P,N and `f_(L)`D. `f_(L)` and F
Answer» Correct Answer - c

Discussion

188.	A body of mass `m_(1) ` moving with uniform velocity of 40 m/s collides with another mass `m_(2)` at rest and then the two together begin to moe wit h uniform velocity of 30 m/s. the ratio of their masses `(m_(1))/(m_(2))` isA. `1:3`B. `3:1`C. `1:1.33`D. `1:0.75`
Answer» Correct Answer - B (b) Initial momentum of the system `m_(1)xx40+m_(2)xx0=40m_(1)` Final momentum of the system=`(m_(1)+m_(2))xx30)` By the law of consevation of momentum, Initial momentum= Final momentum `40m_(1)=(m_(1)+m_(2))30 ` `40 m_(1)-30m_(1)=30m_(2)` `10m_(1)=30m_(2)` `(m_(1))/(m_(2))=3/1`

Discussion

189.	For the arrangement shown in the figure, the reading of spring balance is A. `50 N`B. `100 N`C. `150 N`D. None of the above
Answer» Correct Answer - D `a = ("Net pulling force")/("Total mass") = (10 xx 10 - 5 xx 10)/(10 + 5) = (10)/(3) m//s ^(2)` `5 kg` `T - 5 xx 10 = 5 xx a = (50)/(3)` :. `T = (200)/(3) N` This is also the reading of spring balance.

Discussion

190.	A rod not toching the flor is inserted between two identical boxes (see figure) There is a very small gap between the rod and and boxes ,A horizontal force F is applied at the upper end of the rod There is frication between boxes and the floor Assume that the od remains almost vertical after the force is applied which of hte following is /are correct? A. Both the boxes will move simultaneouslyB. right box will move firstC. left box will move firstD. Friction force on right block is more than left block before the block move
Answer» Correct Answer - b,d

Discussion

191.	A body of mass `m =18kg` is placed on an inclined plane the angle of inclination is `alpha =37^(@)` and is attached to the top end of the slope with a thread which is parallel to the slope. Then the plane slope is moved with a horizontal acceleration of a. Friction is negligible. At what acceleration will the body lose contact with plane .A. Tension in the thread for `a=(5)/(6) m//sec^(2) "is " 12N`B. Tension in the thread for `a=(5)/(6) m//sec^(2) ` is 24 NC. The block loses contact with plane when acceleration becomes`(10)/(3) (m)/(s^(2))`D. The block loses contact with plane when acceleration becomes`(40)/(3) m//s^(2)`
Answer» Correct Answer - a,d

Discussion

192.	A force of `10 N` gives a mass `m_(1)` an acceleration of `10m//s^(2)` and a mass `m_(2)` an acceleration of `20m//s^(2)` What acceleration would it give if both the masses are tid together ?
Answer» Correct Answer - `6.66 m//s^(2)` Here, `m_(1) = (F)/(a_(1)) = (10)/(10) = 1kg` `m_(2) = (F)/(a_(2)) = (10)/(20) = 0.5kg` If a is the acceleration in the combination of masses, then `a= (F)/((m_(1) +m_(2))) = (10)/(1 + 0.5) = 6.66 m//s^(2)` .

Discussion

193.	A small stone of mass 200 g is tied to one end of a string of length 80 cm . Holding the other end in hand , the stone is whirled into a vertical circle What is the minimum speed that needs to be imparted at the lowest point of the circular path , so that the stone is just able to complete the vertical circle ? what would be the tension at the lowest point of circular path ? ` (Take g = 10 m//s^(2)) ` .
Answer» Here ,` m = 200 g = 0.2 kg , ` ` l = r = 80 cm = 0.8 m , upsilon_(L) = ? T_(L) = ? ` From ` upsilon_(L) = sqrt(5 g r ) = sqrt(5 xx 10 xx 0.8 )= 6 .32 ms^(-1) ` ` T_(L) = 6 mg = 6 xx 0.2 xx 10 = 12 N ` .

Discussion

194.	A particle of mass 100 g is moving in a vertical circle of radius 2 m The particle is just looping the loop . What is the speed of the particle and tension in the string at the highest point of the circular path ? . `(g =10 ms^(-2))` .
Answer» Here, ` m = 100 g = 0.1 kg, r= 2 m ` ` upsilon _(H) = ? , T_(H) = ? ` When the particle is just looping the loop ` upsilon _(H) sqrt(g r) = sqrt(10 xx 2) = 4.47 ms^(-1)` At the highest point ` T_(H) = Zero ` .

Discussion

195.	Why does a cyclist lean to one side while going along a curve ? In what direction does he lean ?
Answer» A cyclist leans while going along a curve . By doing so a component of normal reation of the ground is spared to provide him the centripetal force he requires for turning . He has to lean inwards from his vertical position I e towards the center of the circular path .

Discussion

196.	The force on a particle of mass `10g` is `(hati 10+hatj 5)`N If it starts from rest what would be its position at time `t=5s` ?
Answer» Correct Answer - ` vec(r) =(hati 12500 + hatj 6250) m` . From `vec (F) = (hati 10 + hatj5) N,` we have `F_(x) = 10N, F_(y) = 5 N` `:. a xx=(F_(x))/(m)= (10)/(0.01) = 1000 m//s^(2)` As acceleration along x-axis is constant `:. x = u_(x) t + (1)/(2) a_(x) t^(2) = 0 + (1)/(2) xx1000 xx5^(2)` `= 12500m` Similarly `a_(y) = (F_(y))/(m) = (5)/(0.01) = 500m//s^(2)` and `y = u_(y) t + (1)/(2) a_(y) t^(2) = 6250m` Hence position of particle at `t = 5 s` is `vec(r) = (hati 12500 + hatj 6250)m` .

Discussion

197.	Forces of `sqrt 2 N` and `6 sqrt 2 N` are acting on a body of mass 10 kg at an angle of `60^(@)` to each other . Find the acceleration , distance covered and velocity of the body after 10 second , if the body is initially at rest .
Answer» Here ,`F_(1) = 5 sqrt 2 N` `F_(2) =6 sqrt2,0 = 60^(@)` `m= 10 kg , u= 0 , a = ? s = ? upsilon = ?` Resulant force , ` R= sqrt(F_(1)^(2)+ F_(2)^(2) + 2F_(1) F_(2) c` `= sqrt((5sqrt2)^(2) + (6sqrt2)^(2)+2xx5 sqrt2xx6sqrt2xx(1)/2` ` R= sqrt50 +72 + 60 = sqrt182=13.5 N` `a = (R)/m = (13.5)/(10)=1.35 m//s^(2)` `s= ut + (1)/2 at^(2)` `s= 0 + (1)/(2) xx 1.35 xx 10^(2) = 67.5 m` `upsilon = u + "at" = 0 + 1.35 xx 10 = 13.5 m//s` .

Discussion

198.	A dise of mass 10 g is kept floating horizontally by throwing 10 marbles per secound against it from below . If mass of each marble is 5 g Calculate the velocity with which marbles are striking the disc . Assume that marbles strike the disc . Normally and rebound downwards with the same speed.
Answer» Here , ` M = 10 g =10 -^(2) kg , n= 10 , ` ` m= 5 g = 5 xx 10^(-3) kg , upsilon = ` ? To keep the disc floating horizontally Weight of dics = upward force on the disc = rate of change of momentum of marbles ` 10^(-2) xx 9.8 = n xx m xx 2 upsilon = 10 xx 5 xx 10^(-3) xx 2 upsilon ` ` = (upsilon)/(10)= ` ` upsilon = 9.8 xx 10^(-2) xx 10 m//s = 0.98 m//s ` .

Discussion

199.	In the figure given below, with what accelerating does the block of mass m will move? ( Pulley and strings are massless and frictionless)A. `(g)/(3)`B. `(2g)/(5)`C. `(2g)/(3)`D. `(g)/(2)`
Answer» Correct Answer - C ( c) Acceleration `a =[(m_(2)-m_(1))/(m_(2)+m_(1))]g` `=[(5m-m)/(5m+m)]g=(4m)/(6m)g=(2)/(3)g`

Discussion

200.	A rope of mass `0.5kg` is pulling a block of mass `10kg` under the action of force of `31.5N` If the block is resting on a smooth horizontal surface calculate the force of reaction exerted by the block on the rope .
Answer» Correct Answer - `30N` . Total mass `= m + M = 0.5 + 10 = 10.5 kg` `F= 31.5 N` `:.a = (F)/((m + M)) = (31.5)/(10.5) = 3m//s^(2)` Force of reaction exerted block on the rope `R = M xx a = 10 xx 3 = 30 N` .

Discussion

Explore topic-wise InterviewSolutions in .

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