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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.
| 301. |
In the figure shown ,acceraltion of is x (upwards ) Acceleration o fpulleys `p_(3)` with respect to pulley `p_(2)` is y (downwards) and accelarataion of f4 wioth respect to pul,ley `p_(3)` is z (upwards). Then match the following |
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Answer» Correct Answer - (A)s,(B )S |
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| 302. |
What provides centripetal force to a car turning on a level road ? |
| Answer» The force if friction between the tyres and the road provides necessary centripetal force . | |
| 303. |
Assertion : A body can be at rest even when it is under the action of may number of external forces Reason : Because vector sum of the all external forces is zero .A. If both, Assertion and Reason are true and the Reason is the correct explanation of the AssertionB. If both, Assertion and Reason are true but Reason is not a correct explanation of the AssertionC. If Assertion is ture but the Reason is falseD. If both, Assertion and Reason are false |
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Answer» Correct Answer - a When vectore sum of all external forces is zero a body at rest will continue to be at rest . |
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| 304. |
A thief jumps from the upper story of a house with a load on this back. What is the force of the load on his back, when thief is in air ? |
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Answer» Zero . This is because in free fall , g= a ` F = m ( g - a ) = m (g - g ) = Zero ` . |
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| 305. |
Aeroplanes having wings fly at low altitudes while jet planes fly at high altitudes . Way ? |
| Answer» The wings of an aeroplane push the air backwards . The aeroplane noves forward due to reaction of the pushed air At lower altitudes, air is dense . Therefore, The plane receives sufficient reactional push to move forward . In a jet plane , external air is sucked into the plane and compressed Therefore, jet planes fly at high altitudes where air density is small . | |
| 306. |
Consider a vehicle going on a horizontal rod towards east. Neglect any force by the air. The frictional forces on the vehicle by the roadA. is towards east if the vechicle is moving with a unifrom velocityB. is towards east if the vehicle is acceleratingC. must be towards eastD. must be towards west |
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Answer» Correct Answer - b,d When vehicle is moving with unifrom velocity its acc =0 external force applied = 0 Therefore force of friction =0 When vehicle is accelerating towards east force exerted by near wheels on the ground makes force of friction act in the forward direction towards east . |
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| 307. |
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of `12 ms^(-1)` . If the mass of the ball is 0.15 kg , determine the impulse imparted to the ball . (Assume linear motion of the ball). |
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Answer» Here, ` u= 12 ms^(-1) , upsilon = -12 ms^(-1)` , `m= 0.5 kg` , Impulse = ? As Impulse = Change in linear momentum `:.` Impulse `= m (upsilon - u) = 0.15 (-12 - 12)` `= -3.6 Ns` Negative sigh shown that impulse is along negative x - axis , i.e., from the batsman to the bowler. |
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| 308. |
What is the minimum velocity with which a body of mass `m` must enter a vertical loop of radius `R` so that it can complete the loop?A. `sqrt(2 gR`B. `sqrt3gR`C. `sqrt(5gR`D. `sqrtgR` |
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Answer» Correct Answer - c From the knowledge of theory for looping a verticle loop minimum velocity of a body at the lowest point of the loop `=sqrt(5gR)` . |
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| 309. |
A mass of `3kg` descending vertically downwards supports a mass of `2kg` by means of a light string passing over a pulley At the end of `5s` the string breaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.A. 4.9 mB. 9.8mC. 10.6 mD. 2.45 m |
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Answer» Correct Answer - A (a) Acceleration of system before breaking the string was, `a=("Net pulling force")/("Total mass")=(3g-2g)/(5)=(g)/(5)` After 5s velocity of system,`v=at=(g)/(5)xx5=g ms^(-2)` Now, `h=(v^(2))/(2g)=(g^(2))/(2g)=(g)/(2)=4.6m` |
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| 310. |
A body is moving under the action of two force `vec(F_(1)=2hati-5hatj` , `vec(F_(2)=3hati-4hatj` . Its velocity will become uniform under a third force `vec(F_(3)` given by.A. `5hati-hatj`B. `-5hati-hatj`C. `5hati+hatj`D. `-5hati+9hatj` |
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Answer» Correct Answer - D For uniform velocity, acceleration is zero. Hence, resultant force will be zero. `therefore vec(F)_(1)+vec(F)_(2)+vec(F)_(3)=0,` `(2 hati-5hatj)+(3hati-4hatj)+vec(F)_(3)=0 or vec(F)_(3)=(-5hati+9hatj)` |
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| 311. |
A bullet of mass 0.01 kg is fired horizontal into a 4 kg wooden block block at rest , on a horizontal surface. The coefficient of kinetic friction between the block and bullet is 0.25 the combination moves 20 m before coming to rest . With what speed did the bullet strike the block ? |
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Answer» Here , mass of bullet , ` m_(1) = 0.01 kg` mass of wooden block , ` m_(2) = 4 kg ` Coeff . Of kinetic friction , ` mu = 0.25 , ` distance moved by the combination ` s = 20 m ` Let velocity of bullet be `u_(1)`, initial velocity of block `u_(2) = 0` If upsilon is the velocity with which the combination moves , then according to the principle of conservation of linear momentum , `(m_(1) + m_(2) ) upsilon = m_(1) u_(1) + m_(2) u_(2) ` `(0 . 01 + 4 ) upsilon = 0.1 u_(1) + 4 xx 0 ` `upsilon = (0.01u_(1))/(4.01)= (u_(1))/(401)` Force of dynamic friction on the combination ` F = mu R = mu (m_(1) + m_(2)) g` ` = 0 .25 (0.1 + 4 ) xx 9.8 N ` `:.` Retardation produced `a = (f ) /(m_(1)+ m_(2))= (0.25 xx 4.01 xx 9.8 )/(4.01) = 2 .45 ms^(-2)` Using for the combination , the relation `u^(2) - u^(2) = 2 as , 0 - upsilon^(2) = 2 (-2 .45 ) xx 20` `upsilon^(2) = 4.9 xx20 = 98 or upsilon = sqrt(98)` From (i) `u_(1) = 401 upsilon` or `u_(1) = 401 sqrt(98)= 3969 .7 ms^(-1)` . |
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| 312. |
A bullet of mass 10 gram is fired horizontally into a 5 kg wooden block at rest on a horizontal surface The coefficient of kinetic friction between the block and the surface is 0.1 Calculate the speed of the bullet striking the block if the combination moves 20 m before coming rest . |
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Answer» Here, `m_(1) = 10` gram `= 10^(-2) kg ,m_(2) = 5 kg ` `u_(2) = 0 , u_(1) = ? mu_(k) = 0.1 , s = 20 m` Let be the velocity of the combination (bullet and block ) after the bullet pierces the block According to principle of conservation of linear momentum `(m_(1) + m_(2)) upsilon = m_(1) u_(1) + m_(2) u_(2) = m_(1) u_(1)` `upsilon = (m_(1)u_(1))/(m_1 + m_(2))=(10^(-2) u_(1))/(10^(-2) + 5) ` ` upsilon = (u_(1))/(501) ` Now , for the combination `u = upsilon,v = 0,s = 20 m`, `a = -mu_(k) g = - 0.1 xx 9.8 = - 0.98 m//s^(2)` From `upsilon^(2) - u^(2) = 2` as `0 - upsilon^(2) = 2 (-0.98) xx 20 ` `upsilon= sqrt(39.2) = 6 .26 m//s ` From `u_(1) = 501_(upsilon)` ` = 501 xx 6.26 = 3136.26 m//s ` . |
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| 313. |
An impulsive force of 100 N acts on a body for 1 s What is the change in its linear momentum ? |
| Answer» Change in linear momentum = Impulse ` = F xx t = 100 xx 1 = 100 Ns ` . | |
| 314. |
Can a single isolated force exist in nature ? |
| Answer» No this would violate Newton s #rd law of motion . | |
| 315. |
In an isolated system……of……is…..and is not affected by . |
| Answer» vector sum, linear momenta of any number of bodies , constant , their mutual action and reaction . | |
| 316. |
Burnt out speed of a rocket …..when…… . |
| Answer» is the speed of the rocket , its entire fuel is burnt up . | |
| 317. |
A ship of mass `3xx10^7kg` initially at rest, is pulled by a force of `5xx10^5N` through a distance of 3m. Assuming that the resistance due to water is negligible, the speed of the ship is |
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Answer» Correct Answer - `0.1 ms^(-1)` Here, `m=3 xx 10^(7)kg, u=0, s=3m` `F =5 xx 10^(4) N, upsilon = ?` As `a = F//m = (5 xx 10^(4))/(3 xx 10^(7)) = (5)/(3) xx10^(-3) m//s^(2)` From `v^(2) - u^(2) = 2 as ` `v^(2) - 0 = 2 xx (5)/(3) xx 10^(-3) xx3 =10^(-2)` `v= 10^(-1) ms^(-1) = 0.1 ms^(-1)` . |
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| 318. |
A ship of mass `3xx10^7kg` initially at rest, is pulled by a force of `5xx10^5N` through a distance of 3m. Assuming that the resistance due to water is negligible, the speed of the ship isA. `1.5m//sec`B. `60m//sec`C. `0.1m//sec`D. `5m//sec` |
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Answer» Correct Answer - C `F=ma` `implies a=F/m=(5xx10^4)/(3xx10^7)=5/3xx10^-3ms^-2` Also, `v^2-u^2=2as` `impliesv^2-0^2=2xx5/3xx10^-3xx3=10^2` `impliesv=0.1ms^-1` |
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| 319. |
A parabolic bow1 with its bottom at origin has the shape `y = (x^(2))/(20)` where `x` and `y` are in metre The maximum height at which a small mass m can be placed on the bowl without slipping is (coeff of static friction `0.5` .A. `1.25Nm`B. `2.5m`C. `1.0m`D. `4.0m` |
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Answer» Correct Answer - a As `y =(x^(2))/(20) , (dy)/(dx) =(2x)/(20)` Angle of repose `theta_(r)` is such that `mu = tan theta_(r) = (dy)/(dx) = (2x)/(20) = (x)/(10)` `x = 10 mu = 10 xx 0.5 = 5m` Corresponding value of `y = (x^(2))/(20) = (5^(2))/(20) = 1.25m` . |
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| 320. |
A parabolic bow1 with its bottom at origin has the shape `y = (x^(2))/(20)` where `x` and `y` are in metre The maximum height at which a small mass m can be placed on the bowl without slipping is (coeff of static friction `0.5` .A. `2.5 m`B. `1.25 m`C. `1.0 m`D. `4.0 m` |
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Answer» Correct Answer - B |
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| 321. |
A metre scale is moving with uniform velocity. This implies .A. the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.B. he force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.C. the total force acting on it need not be zero but the torque on it is zero.D. neither the force nor the torque need to be zero. |
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Answer» Correct Answer - B When a metre scale is moving with uniform velocity, the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero. |
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| 322. |
A mater scale is moving with unifrom velocity This implies .A. the force acting on the scale is zero but a torque about the center of mass can act on the scale .B. the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero .C. the total force acting on it need not be zero but the torque on it is zero .D. neither the force nor the torque need to be zero. |
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Answer» Correct Answer - b When a metre scale is moving with unifrom velocity the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero . |
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| 323. |
Calculate the ratio `m_(0)//m` for a rocket to attain the escape velocity of `11.2km s^(-1)` after starting from rest, when maximum exhaust velocity of gases is `1.6km//s` . |
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Answer» Correct Answer - `1096` . Here, `upsilon = 11.2 km//s, u = 1.6 km//s ((m_(0))/(m))= ?` From `upsilon = u log_(e) ((m_(0))/(m))` `= 2.3026u log_(10) ((m_(0))/(m))` `log_(10) ((m_(0))/(m)) = (upsilon)/(2.3026 xxu) = (11.2)/(2.3026 xx 1.6) = 3.0400` `:. ((m_(0))/(m)) = antilog 3.0400 = 1096` . |
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| 324. |
Mark the correct statements .A. The electromagnetic force between two protons is always greater than the gravitational force between themB. The nuclear force between two protons is always greater than the electromagnetic force between themC. The gravitational force between two protons may be greater than the nuclear force between themD. Electromagnetic force between two protons may be greater than the nuclear force acting between them |
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Answer» Correct Answer - a,b Nuclear force gt electromagnetic force gravitational force between protons . |
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| 325. |
Figure shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region A. `AB`B. `BC`C. `CD`D. `DE` |
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Answer» Correct Answer - a,c Displaement time graph `AB` and `CD` are linear Therefore vel is uniform acc `=0.F =0` . |
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| 326. |
A cricket ball of mass 150 kg is moving with a velocity of `12 m//s` and is hit by a bat so that ball is turned back with a velocity of `20 m//s` . The force of the blow acts for 0.01 s on the ball . Find the average force exerted by the bat on the ball. |
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Answer» Here ,` m = 150 g = 0.15 kg = 0.15 kg ` ` u =12 m//s, upsilon = -20 m//s , t =0.01 s , F = ? ` As `Fxx 1` = impulse = change in momentum ` F xx 0.01 = m (upsilon - u) = 0.150 (-20 - 12) ` ` F = (-0.150 xx 32 )/(0.01)= - 480 N ` . |
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| 327. |
According to Newton s second law of motion `F =ma` where `F` is the force required to produce an acceleration a in a body of mass m If a = 0 then `F = 0` no external force is required to move a body uniformly along a straight line If a force `F` acts on a body for t seconds the effect of the force is given by Impulse =F `xx` t =change in linear momentum of the body Average force exerted by the bat is .A. `480N`B. `120N`C. `1200N`D. `840N` |
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Answer» Correct Answer - a As Impulse `=F xx t` `:. F = ("Impulse")/(t) = 4.80N` . |
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| 328. |
Impulse of a force is …….and…… . |
| Answer» product of force , time for which the force acts. | |
| 329. |
One newton force is ……..produce an acceleration of…….in a body of……. . |
| Answer» that much force which ,` 1ms^(2)` , mass 1 kg . | |
| 330. |
A soild block of mass 2 kg is resting inside a cube as shown in the figure .The cube is moving with a veocity `v=5thati +2hatj m//s` Here , t is is time in second the block is at rest with respect to the cube and cefficient of frision between the surface of ube and block is 0.6 then ,[take `g=10m//s^(2)`] A. force of friction acting on the block is 10 NB. force of friction acting on the block is 4 NC. the total force exerted by the block on the cube is `10sqrt(5) N`D. a,c |
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Answer» Correct Answer - a,c |
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| 331. |
A `2 kg` block is present against a rough wall by a force `F = 20 N` as shown in figure .Find acceleration of the block and force of friction acting on it .`(Take g = 10 m//s^(2))` |
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Answer» `N = 20 N=` applied force `mu_(s) N = 16 N` and `mu_(k) N = 12 N` Driving force `F = mg` (downward) Since `F gt mu_(k) N`, block will slide downwards and kinetic frictioin of `12 N` will act in upward direction `:. a=(F - mu_(k) N)/(m)` (downward) `= (20 - 12)/(2) = 4 m//s^(2)` (downward) |
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| 332. |
In the show arrangement mass of A=1 kg, mass of B= 2kg. Coefficient of friction between A and B =0.2 There is no friction between B and ground. The frictional force exerted by A on B equals.A. 2 NB. 3 NC. 4 ND. 5 N |
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Answer» Correct Answer - A (a) Block A moves due to friction. Maximum acceleration of A can be `(f_("max"))/(m) " or " (mu mg)/(m)` or `mu g=0.2xx10=2 ms^(-2)`. If both the blocks move together, then combined acceleration of A and B can be `(10)/(3)=3.33 ms^(-2)`. Since, this is more than the maximum acceleration of A . Slipping between them will take place and force of friction will be maximum of `mu m_(A)g=2N`. |
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| 333. |
A block of mass m is placed on the top of another block of mass M as shown in the figure. The coefficient of friction between them is `mu`. The maximum acceleration with which the block M may move so that m also moves along with it isA. `mu g`B. `mu(M)/(m) g`C. `mu(m)/(M) g`D. `(g)/(mu)` |
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Answer» Correct Answer - A (a) m will move by friction, `f_("max")=mu mg implies(a_(m))_("max")=(mu mg)/(m)=mu g` |
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| 334. |
A 10 N force is applied on a body produces an acceleration of 1 `m//s^(2)` . The mass of the body isA. 5 kgB. 10 kgC. 15 kgD. 20 kg |
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Answer» Correct Answer - B |
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| 335. |
A balloon has a mass of 10 gram in air. The air escapes from the balloon at a unifrom rate with a velocity of `5 cm //s` and the balloon shrinks completely in 2.5 s . Calculate the average force acting on the balloon. |
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Answer» Here , ` m = 10 gram , upsilon = 5 cm//s `, ` (dm)/(dt)= (10)/(2.5)= 4 gram //sec . , F= ? ` ` F = (dp)/(dt)=(d)/(dt) (m upsilon)=((dm)/(dt)) upsilon = 4 xx 5 ` ` 20 "dyne"` |
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| 336. |
In moving a body of mass m once up and down a smooth incline `0` total work done is (S is length of the plane).A. `mg sin theta xx S`B. `mg cos theta = S`C. `2 mu mg cos theta xx s`D. zero |
| Answer» Correct Answer - c | |
| 337. |
A ball rolling on ice with a velocity of `4 .9 m//s` stops after travelling 4 m. If `g = 9.8 m//s^(2)` what is the coefficient of friction ?A. `0.1`B. `0.2`C. `0.3`D. `0.4` |
| Answer» Correct Answer - b | |
| 338. |
A ball rolling on ice with a velocity of `4 .9 m//s` stops after travelling 4 m . If `g = 9.8 m//s^(2)` what is the coefficient of friction ? |
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Answer» Here , ` u = 4.9 m//s , upsilon = 0 , s = 4 m ` ` g = 9.8 m//s^(2) , mu = ` ? ` upsilon^(2) - u^(2) = 2 as = 2 (-mu g ) s ` ` mu= (upsilon^(2) - u^(2))/(-2gs) = (0-(4.9)^(2))/(-2 xx 9.8 xx 4 )= 0.31 ` . |
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| 339. |
The `10 kg` block is moving to the left with a speed of `1.2 m//s` at time `t = 0` A force `F` is applied kinetic friction is `mu_(k) = 0.2`. Determine the time `t` at which the block comes to a slip. `(g = 10 m//s^(2))` |
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Answer» `mu_(k) mg = 0.2 xx 10 xx 10 = 20 N` For `t le 0.2 s` Retardation ` a_(1) = (F + mu_(k) mg)/(m)` `= (20 + 20)/(10) = 4 m//s^(2)` At the end of `0.2 s` `v = 1.2 - 4 xx 0.2 = 0.4m//s` For `t gt 0.2 s` Retardation `a_(2) = (10 + 20)/(10) = 3 m//s^(2)` Block will come to rest after time `t_(0) = (v)/(a_(2)) = (0.4)/(3) = 0.13 s` `:.` Total time `= 0.2 + 0.13 = 0.33 s` |
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| 340. |
In the question number 82, the maximum permissible speed to avoid slipping isA. `18.6"ms"^(-1)`B. `28.6"ms"^(-1)`C. `38.6"ms"^(-1)`D. `48.6"ms"^(-1)` |
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Answer» Correct Answer - C The maximum permissible speed is given by `v("max")=sqrt((Rg+(mu+tan theta))/(1-mutantheta))=sqrt((300xx10xx(0.2+0.27))/(1-0.2xx0.27))` `=38.6ms ^(-1)` |
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| 341. |
A car travels on a flat, circular track of radius `200 m` at `30 ms^(-1)` and has a centripetal acceleration `= 4 .5 ms^(-2)`. (a) If the mass of the car is 1000 kg , what frictional force is required to provide the acceleration ? (b) If the coefficient of static friction is 0.8, what is the maximum speed at which the car can circle the track ? |
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Answer» Here , ` r = 200 m , upsilon = 30 ms -1 ` Centripetal acc , `a = 4.5 ms^(-2) , m = 1000 kg ` Frictional force required ` F = ? ` As F = Accelerating force `= m a = 1000 xx 4.5 ` `= 4500 N` (b) `mu = 0 .8 , upsilon = ?` `upsilon = sqrt(mu r g)= sqrt(0.8 xx 200 xx 9.8 )=39.6 ms^(-1)` . |
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| 342. |
Out of stiatic friction limiting friction and dynamic friction which is largest ? |
| Answer» Limiting friction is largest of all . | |
| 343. |
A soda water bottele is falling freely . Will the bubbles of gas rise in the water of the bottle ? |
| Answer» Bubble will not rise in water . This is because water in freely falling bottle is in the state of weightlessness. No upthrust acts on the bubbles . | |
| 344. |
Which of the following is a self adjusting force ?A. kinetic frictionB. limiting frictionC. static frictionD. all the three |
| Answer» Correct Answer - c | |
| 345. |
A block of mass m is attached with a massless in string. Breaking strength of string is 4mg. Block is moving up. The maximum acceleration and maximum retardation of the block can be A. 4g,3gB. 4g,gC. 3g.gD. 3g.4g |
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Answer» Correct Answer - c |
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| 346. |
A car accelerates on a horizontal road due to the force exerted byA. the engine of the carB. the driver of the carC. the car on earthD. the road on the car |
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Answer» Correct Answer - D A car acceleration on a horizontal road due to the force exerted by the road on the car. |
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| 347. |
A stone of mass 5 kg is tied to a string of length 10 m is whirled round in a horizontal circle. What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?A. `10 "ms"^(-1)`B. `15 "ms"^(-1)`C. `20 "ms"^(-1)`D. `25 "ms"^(-1)` |
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Answer» Correct Answer - C Here, m=5kg, r=10m, `T_("max")=200N` `As T_(max)=(mv_("max)^(2))/(r)` `therefore v_(max)^(2)=(T_("max")xxr)/(m)` `therefore v_("max")^(2)=(200xx10)/(5)=400 Rightarrow v_("max")=20ms^(-1)` |
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| 348. |
A body of mass 10 kg is placed on an inclined surface of angle `30^(@)` . If coefficient of limiting friction is `1 //sqrt3` , find the inclined plane. Force is being exerted parallel to the inclined plane . |
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Answer» Here , ` m = 10 kg , theta = 30^(@) , mu = (1)/sqrt3 ` As is clear from force required just to push the body up the inclined plane is ` F = mg sin theta + f ` ` = mg sin theta + mu R ` ` = mg sin theta + mu mg cos theta ` `= mg (sin theta + mu cos theta) ` ` = 10 xx 9.8 (sin 30^(@) + (1)/sqrt(3 ) cos 30^(@)) ` ` F = 98 (0.5 + (1)/(sqrt3) xx sqrt(3)/(2)) = 98 N ` . |
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| 349. |
A stone of mass 4 kg is attached to a string of 10 m length and is whirled in a horizontal circle . Calculate the max . Velocity with which the stone can be whirled if the string can withstand a maximum tension of 160 N . |
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Answer» Here, ` m = 4 kg , r = 10 m ` ` upsilon = ? T = F = 160 N ` ` T = F =( mu upsilon^2)/(r) , upsilon = sqrt((Fxxr)/(m)) = sqrt((160 xx 10)/(4)) ` ` = 20 m//s ` . |
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| 350. |
Consider the system shown in the figure. The wall is smmoth, but the surfaces of blocks A and B in contact are rough. The friction on B due to A in equilibrium is A. upwardB. downwardC. not definedD. the system cannot remain in equilibrium. |
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Answer» Correct Answer - d |
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