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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.
| 1151. |
A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration `alpha` in the y-direction. Its equation of motion is `y= betax^2`. Its velocity component in the x-direction isA. variableB. `sqrt(((2a)/(b)))`C. `(a)/(2b)`D. `sqrt(((a)/(2b)))` |
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Answer» Correct Answer - D |
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| 1152. |
A particle of mass `m` moves on the ` x- axis ` as follows : it starts from rest at ` t = 0`, from the point `x = 0`, and comes to rest at ` t = l` at the point `x = 1`. No other information is available about its motion at intermediate times `( 0 lt t lt l)` . If `alpha ` denotes the instantaneous accelartion of the particle , then :A. `alpha` cannot remain positive for all t in the interval `o le t le 1`B. `|alpha|` cannot exceed 2 at any point in its pathC. `|alpha|` must be `ge 4` at some point or points in its pathD. `alpha` must change sign during the motion, but no other assertion can be made with the information given |
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Answer» Correct Answer - A::C |
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| 1153. |
Blocks A and B have masses of 2 kg and 3 kg respectively. The ground is smooth. P is an external force of 10 N. The force exerted by B on A is (a) 4 N (b) 6 N (c) 8 N (d) 10 N |
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Answer» Correct answer is: (b) 6 N Find the acceleration of the two-body system. Then make a free-body diagram of any one unit. Apply F = ma to that unit. |
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| 1154. |
A particle starts from the origin at t = 0 and moves in the xy plane with constant acceleration a in the y-direction. Its equation of motion is y = bx 2 . The x-component of its velocity is(a) variable(b) √(2a/b)(c) a/2b(d) √(a/2b) |
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Answer» Correct Answer is:(d) √(a/2b) y = bx2, y = 2bx x, y = a = 2bx2 + 2bxx, Here, x = 0, ∴ x = √(a/2v) |
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| 1155. |
A man slides down a light rope whose breaking strength is η times his weight (η < 1). What should be his maximum acceleration so that the rope just breaks?(a) ηg(b) g(1 - η) (c) g / 1 + η(d) g / 2 - η |
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Answer» Correct answer is: (b) g(1 - η) |
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| 1156. |
The blocks B and C in the figure have mass m each. The strings AB and BC are light, having tensions `T_(1)` and `T_(2)` respectively. The system is in equilibrium with a constant horizontal force mg acting on C.A. `tan theta_(1) = 1//2`B. `tan theta_(2) = 1`C. `T_(1) = sqrt(5)mg`D. `T_(2) = sqrt(2)mg` |
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Answer» Correct Answer - A::B::C::D |
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| 1157. |
A monkey of mass m kg slides down a light rope attached to a fixed spring balance, with an acceleration a. The reading of the spring balance is W kg. [g = acceleration due to gravity](a) The force of friction exerted by the rope on the monkey is m(g - a) N.(b) m = Wg / g - a(c) m = W ( 1+ a/g)(d) The tension in the rope is Wg N. |
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Answer» Correct Answer is: (a, b, & d) |
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| 1158. |
Aspring of weight W and force constant k is suspended in a horizontal position by two light strings attached to its two ends. Each string makes and angle `theta` with the vertical .The extension of the sporing isA. `(W//4k)tantheta`B. `(W//2k)tantheta`C. `(W//4k)sintheta`D. 0 |
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Answer» Correct Answer - B |
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| 1159. |
Two trains A and B are approaching each other on a straight track, the former with a uniform velocity of 25 m/s and other with 15m/s, when they are 225 m a part brakes are simultaneously applied to both of them. The deceleration given by the brakes to thetrain B increases linearly with time by `0.3 m//s^(2)` every second, while the train A is given a uniform deceleration, (a)What must be the minimum deceleration of the train A so that the trains do not collide ? (b) What is the time taken by the trains to come to stop ?A. `5s`B. `25s`C. `15s`D. `10s` |
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Answer» Correct Answer - D |
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| 1160. |
Two trains A and B are approaching each other on a straight track, the former with a uniform velocity of 25 m/s and other with 15m/s, when they are 225 m a part brakes are simultaneously applied to both of them. The deceleration given by the brakes to thetrain B increases linearly with time by `0.3 m//s^(2)` every second, while the train A is given a uniform deceleration, (a)What must be the minimum deceleration of the train A so that the trains do not collide ? (b) What is the time taken by the trains to come to stop ?A. `5m//s^(2)`B. `2.5m//s^(2)`C. `1.5m//s^(2)`D. `7.5m//s^(2)` |
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Answer» Correct Answer - B |
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| 1161. |
Two trains A and B are approaching each other on a straight track, the former with a uniform velocity of 25 m/s and other with 15m/s, when they are 225 m a part brakes are simultaneously applied to both of them. The deceleration given by the brakes to thetrain B increases linearly with time by `0.3 m//s^(2)` every second, while the train A is given a uniform deceleration, (a)What must be the minimum deceleration of the train A so that the trains do not collide ? (b) What is the time taken by the trains to come to stop ? |
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Answer» Correct Answer - `[2.5 m//s^(2), 10.0s]` |
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| 1162. |
A spring of weight W and force constant k is suspended in a horizontal position by two light strings attached to its two ends. Each string makes an angle θ with the vertical. The extension of the spring is (a) (W/4k)tan θ(b) (W/2k)tan θ(c) (W/4k)sin θ (d) 0 |
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Answer» Correct Answer is: (b) (W/2k)tan θ Let T = tension in each string. 2Tcos θ = W and Tsin θ = kx, x = extension of spring or (W/2cos θ) sin θ = kx or x = w tan /2k. |
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| 1163. |
In the arrangement shown, the pulleys are fixed and ideal, the strings are light, `m_(1)gtm_(2)` and S is a spring balance which is itself massless. The readings of S (in units of mass) is A. `m_(1) - m_(2)`B. `(1)/(2)(m_(1)+m_(2))`C. `(m_(1)m_(2))/(m_(1)+m_(2))`D. `(2m_(1)m_(2))/(m_(1)+m_(2))` |
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Answer» Correct Answer - D |
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| 1164. |
A football is kicked at a speed of `20 m//s` a projection angle of `45^(@)`. A receiver on the goal line ` 25` metres away in the direction to the kink runs the same instant to meet the ball. Befrom it hits the ground? |
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Answer» Here, u=20 m//s, `theta =45^(@), d=25 m`. Herizontal range , `R=u^(2)/g sin 2 theta = (20^(2))/(9.8) sin 2 xx 45^(@) = 40.82 m` Time of flight , `T =(2 u sin theta)/g =(20^(2))/99.8) xx sin 45^(@)` `=2.886 s ` Since the goal man is already `25 m` away in the direction of the ball so to catch the ball, he is to cover a distane `=40. 82 -25 =15.92 m`, in time `2. 886 s`. Therfore the velocity of goal manto catch the ball is ` v= (15.82)/(2.886) =5. 481 ms^(-`). |
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| 1165. |
A projectile of mass `2kg` has velocities `3m//s` and `4m/s` at two points during its flight in the uniform gravitational field of the earth. If these two velocities are perpendicular to each other, then the minimum kinetic enerty of the particle during its flight isA. `6.32J`B. `8.40J`C. `16.32J`D. `5.76J` |
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Answer» Correct Answer - D |
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| 1166. |
A ball is thrown form a point on a ground at some angle of projection. At the same time a bird starts form a point directly above this point of projection at a height h horizontally with speed `u`. Given that in its flight ball just touches the bird at one point. Find the distance on ground where ball strikesA. `2usqrt((h)/(g))`B. `usqrt((2h)/(g))`C. `2usqrt((2h)/(g))`D. `usqrt((h)/(g))` |
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Answer» Correct Answer - C `v_(x) = u, h = (v_(y)^(2))/(2g), v_(y) = sqrt(2gh)` `R = (2V_(x)V_(y))/(g) rArr R = (2usqrt(2gh))/(g) rArr R = 2u sqrt((2h)/(g))` |
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| 1167. |
STATEMENT-1: The avergae speed the average velocity of the aprticles are different physical quantities. STATEMENT-2: The avergae speed and average velocity have same dimensions.A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - D | |
| 1168. |
Displacement-time equation of a particle moving along x-axis is `x=20+t^3-12t` (SI units) (a) Find, position and velocity of particle at time t=0. (b) State whether the motion is uniformly accelerated or not. (c) Find position of particle when velocity of particle is zero. |
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Answer» Correct Answer - A (a) `x=20+t^3-12t` …(i) At t=0, `x=20+0-0=20m` Velocity of particle at time t can be obtained by differentiating Eq. (i) w.r.t. time i.e. `v=dx/dt=3t^2-12` …(ii) At t=0, `v=0-12 =-12m//s` (b) Differentiating Eq. (ii) w.r.t. time t, we get the acceleration `a=(dv)/(dt) =6t` As acceleration is a function of time, the motion is non-uniformly accelerated. (c) Substituting v=0 in Eq. (ii), we have `0=3t^2-12` Positive value of t comes out to be 2s from this equation. Substituting `t=2s` in Eq. (i), we have `x=20+(2)^3-12(2)` or `x=4m` |
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| 1169. |
A body covered a distance of `l` metre along a semicircular path. Calculate the magnitude of displacement of the body, and the ratio of distance to displacement. |
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Answer» Let (r ) be the radius of semicircular path. Here, ` l=(2 pi r//2)` or ` r =l//pi`. Diameter `=2 r =2 l// pi , Magnitude of displacement `= diametr `=2 l//pi` (Distance )/(Displacemnt) = l/(2// pi) =(pi)/2`. |
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| 1170. |
Usually average speed means the rationof total distace travelled to the total time elapsed. However, sometimes thephrase ` average speed can vean the magnitude of the average velocity. Are the two same ? Discuss. |
| Answer» It is not correct to say that the average speed of the body is ewual to the magnitude of the average velocity because they have got dufferbt meanings. As average sped, v_(av)= total distance travelled//time taken and magnitude of average velocity `= | vec v_(av) | =| displacement//time |`. As, the distance travelled by a moving body is always positive and can never be zero or negative , whereas the displacement of body can be wero, negative or positiove. So, distance `ge` displacenent. gtbrgt Hence, ` v_(av) ge | vec v_(av) |` . | |
| 1171. |
A ball is projected with velocity (u) at an anle ` alpha` with horizontal plane. What is its speed when it makes an angle ` theta` whith the horizontal plane ? |
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Answer» Let (v) be the velocityoa ball at an instant, when it makes and angle ` beta` with the horizontal. The horizontal component velocity of the ball ` = v cos beta` Initial horizontal component velocity of ball ` = v cos beta` In angular grojection of a projectile, the horizontal component velocity remains unchanged, hence `v cos beta =u cos alpha` or `v=u cos alpha //cos beta`. |
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| 1172. |
The graph as shown in. below descrines the motion of a ball rebounding from a horizontal surface being released from a point above the surface. Assume that the ball colledes each time with the floor inelastically. The quantity represented on the y-axis in the is the ball`s (take upward direction as positive) .A. DisplacementB. VelocityC. AccelerationD. Momentum |
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Answer» Correct Answer - A We know that for a body throun up, its displacement is gimen as `S=ut -(1)/(2) g t^(2)`. So the `s-t` graph is parabolice downwards, Also the ball collides inelastically, so it will revound to hass height every time as shown in the graph `t_(1), t_(2)`, `t_(3)`, are the instants when the ball colledes with ground, Here slope of the `s-t` graph is suddenly changing from negatibe (downwrds) and agter collision is positive (upwards). |
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| 1173. |
The velocity -time of a body falling from rest under gravity and rebounding from a solid surface is represented by which of the following graphs?A. B. C. D. |
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Answer» Correct Answer - A Initially velocity increases downwards (negative) and after rebound it becomes positive and then speed us decreasing due to acceleration of gravoty `(darr)` |
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| 1174. |
The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]` Range OA is :-A. `sqrt(3)/2`B. `sqrt(3)/4`C. `sqrt(3)`D. `3/8` |
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Answer» Correct Answer - A Range of A `=(u^(2) sin2theta)/g=(10xxsin 120^(@))/10=sqrt(3)/2` |
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| 1175. |
The graph between the displacement `x` and time `t` for a particle moving in a straight line is shown in the figure. During the interval `OA, AB, BC` and `CD` the acceleration of the particle is `OA, AB, BC, CD` A. `OA=+, AB=0, BC=+, CD=+`B. `OA=-, AB=0, BC=+, CD=0`C. `OA=+, AB=0, BC=-, CD=+`D. `OA=-, AB=0, BC=0, CD=+` |
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Answer» Correct Answer - D `a=(d^(2)x)/(dt^(2))=` change in velocity w.r.t the time For `OA rarr` velocity decreases so a is negative For `AB rarr` velocity constant so a is zero. For `BC rarr` velocity constant so a is zero For `CD rarr` velocity increases so a is positive. |
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| 1176. |
Displacement-time graph of a particle moving in a straight line is as shown in figure. (a) Find the sign of velocity in regions oa,ab, bc and cd. (b) Find the sign of acceleration in the above region. |
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Answer» Correct Answer - A::B (a) Velocity=slope of s-t graph `:.` Sign of velocity=sign of slope s-t graph (b) Let us discus any one of them, let the portion cd, slope of s-t graph (=velocity) of this region is negative but increasing in magnitude. Therefore sign of velocity and acceleration both are negative. |
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| 1177. |
In a projectile motion let `t_(OA)=t_(1)` and `t_(AB)=t_(2)`.The horizontal displacement from `O` to `A` is `R_(1)` and from `A` to `B` is `R_(2)`.Maximum height is `H` and time of flight is `T`.If air drag is to be considered, then choose the correct alternative(s). A. `t_(1)` will decrease while `t_(2)` will increaseB. H will increaseC. `R_(1)` will decrease while `R_(2)` will increaseD. T may increase or decrease |
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Answer» Correct Answer - A::D As air drag reduces the vertical component of velocity so time to reach maximum height will decrease and it will decrease the downward vertical velocity hence time to fall on earth increases. |
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| 1178. |
Discuss about the Principles of Physical Independence of Motions. |
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Answer» (1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other.This is called the principle of physical independence of motions. (2) The velocity of the particle can be resolved into two mutually perpendicular components. Horizontal component and vertical component. (3) The horizontal component remains unchanged throughout the flight. The force of gravity continuously affects the vertical component. (4) The horizontal motion is a uniform motion and the vertical motion is a uniformly accelerated retarded motion. |
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| 1179. |
What is projectile. |
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Answer» A body which is in flight through the atmosphere but is not being propelled by any fuel is called projectile. |
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| 1180. |
List the Assumptions of Projectile Motion. |
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Answer» (1) There is no resistance due to air. (2) The effect due to curvature of earth is negligible. (3) The effect due to rotation of earth is negligible. (4) For all points of the trajectory, the acceleration due to gravity ‘g’ is constant in magnitude and direction. |
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| 1181. |
A bird is tossing (flying to and fro 0 between two cars moving towards each other on a straight road. On car has a speed of `18 km //h` while the other has the speed of ` 27 km //h`. The bird starts moving from first caer towards the other and is moving with the speed of ` 36 km //h` and when the two cars were separated by ` 36 km`. What is the total distance comered by the bird ? What is the total displacement of the bird ? |
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Answer» Relative velocity of each car w.r.t. another car ` =27 + 27 =54 km //h` Relative velocity of bird w.r.t. a car approaching bird ` =18 + 27 =45 km //h` Time to meet the bird with approaching car, ` t= (36 km ) /(45 km//h) = 4/5 =0. 8 h` Total distance covered by bird ` =v-b t= 36 xx 0.s =28 .8 km` . |
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| 1182. |
Two towns `A` and `B` are connected by a regular bus service with a bus leaving in either direction every `T` min `A` man cycling with a speed of `20 km h^(-1)` in the direction `A` to `B` notices that a bus goes past him every `18 min` in the direction of his motion, and every `6 min` in the opposite direction. What is the period `T` of the bus service and with what speed (assumed constant )do the buses ply on the road? |
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Answer» Let speed of each bus `=v hm h^(-1)` The distance between the nearest buses plying on either car `=vT km` (i) For buses going grom town `A` to `B`: Relative speed of bus in this direction go past the cyclist after every `18 min`. Theteftore, separation between the buses `=(v-20)xx(18)/(60)` From (i), `(v-20)xx(18)/(60)=vT` (ii) For buses coming from `B` to `A` : The relative velocity of bus respect to man`=(v+20)` `becuase (v+20)xx(6)/(60)=vT` ...(iii) Solving (ii) and (iii), we get `v=40 km h^(-1)` and `T=(2)/(30)h`. |
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| 1183. |
Two towns `A` and `B` are connected by a regular bus service with a bus leaving in either direction every `T` min. `A` man cycling with a speed of `20 km h^(-1)` in the direction `A` to `B` notices that a bus goes past him every `18 min` in the direction of his motion, and every `6 min` in the opposite direction. What is the period `T` of the bus service and with what speed (assumed constant )do the buses ply on the road? |
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Answer» Let ` km h^(-1)` be the constant speed with which the buses pky between the torwns ` A and B` The relative velocity of the bus ( for the motion ` A to B` 0 with respect to the cylist ( i.e. in the dierction in which the cyclist is going = 9v - 20 0 km h^(-1)`. The relative velocity of the bus from ` B to A` with cyclist ` = ( v+ 20 0 km h^(-1)` The distanc etravelled by the bus in time ` T ( minutes) = vT` As per question ` ( vT)/(v- 20) = 18` or ` v T = 18 v- 18 xx 20 ` ...(i) and ` ( vt) /( v+ 20) = 6 or ` v T = 6 v + 20 xx 6` (ii) Equating (i) & (ii) we get ` 18 v- 18 xx 20 = 6 v+ 20 xx 6` or ` 12v = 20 6 + 18 xx 20 = 480 or v= 40 km h^(-1_` Puttig this value of (v) in(i) we get ` 40 T= 18 xx 40 - 18 xx 20 = 18 xx 20 or ` T = 18 xx 20// 40 = 9 min` |
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| 1184. |
On a two lane road , car (A) is travelling with a speed of `36 km h^(-1)`. Tho car ` B and C` approach car (A) in opposite directions with a speed of ` 54 km h^(-1)` each . At a certain instant , when the distance (AB) is equal to (AC), both being ` km, (B) decides to overtake ` A before C does , What minimum accelration of car (B) is required to avoid and accident. |
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Answer» Velocity of car ` A= 36 km h^(-1) = 10 ms^(-1)` , Velocity oa cr ` A ot C= 54 km h^(-1) = 15 ms^(-1)` , Relative velocity of ` B w.r.t. A =15 ms^(-1)` , Relative velocity of ` C w.r.t. A = 15 + 10 = 25 ms^(-10` As, ` AB =AC = 1 km = 1000 m` Time available to ` B or C` for crossing ` A = ( 1000)/(25) = 40 s` If car ` (B) accelrates with accleration ` a, to cross (A) bofore car (C ) does. then ` u= 5 ms^(-1) , t= 40 s, S= 1000 m , a= ?` Using , `S = ut + 1/2 at^2 `, we have `1000 = 5xx 40 + 1/2 xx a xx 40^@ or 1000 - 200 = 800 a or a =1 s//s^2` |
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| 1185. |
A fly wheel is making ` 300` rpm. Its angular velocity in radian per second is . |
| Answer» Correct Answer - (c ) | |
| 1186. |
A bullet is fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet can be :A. `15^(@)`B. `30^(@)`C. `60^(@)`D. `75^(@)` |
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Answer» Correct Answer - A |
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| 1187. |
Ball are dropped from the roop fo towar at fived interval if tiem . At the moment when ` 9th ball reaches the groun the nth ball is (3//4) the heith of the tower Wgat the vale of ` n ? G= 10 m//s^2`. |
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Answer» Let (t) be the time interval between two successive balls while falling . When `9th ` ball reaches the ground, the Ist ball is just to be dropped, os the time taken by the 9th ball to reach the groun ` =(9 -1) t= 8 t` . `:. S= 1/2 xx g xx g xx (8t)^2 = 1/2 10 xx 64t^2 = 320 t^2 ` ...(i)` Time taken by nth ball to fall `= (n-1) t` Distance travelled ` = s- 3 S/4` :. ` S/4 = 1/2 xx [(n-1)^2 t]^2 = 5 (n-a)t` or S= 20 (n-1)^2 t^2` ...(ii) From (i) and (ii) ` 320 t^2 = 20 (n-1)^2t^2` On solving, ` n=5`. |
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| 1188. |
A river is flowing with velocity `5km//hr` as shown in the figure. A boat start from A and reaches the other bank by covering shortest possible distance. Velocity of boat in still water is `3km//hr`. The distance boat covers is. A. `500 m`B. `400 sqrt(2)m`C. `300 sqrt(2) m`D. `600 m` |
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Answer» Correct Answer - A When `v_(r) gt v_(br)` `{:("for minimum drift",,v_(r)=5 km//hr),(,,v_(br)=3 km//hr),(sin theta=v_(br)/v_(r)=3/5,,b=300 m=0.3 km):}` `theta=37^(@)` velocity of boat along x-axis so `v_(x)=v_(r)-v_(br) sin theta` `v_(x)=5/1-3xx3/5=16/5 km//hr` velocity of boat along y-axis `v_(y)=v_(br) cos theta=3xx4/5=12/5 km//hr` Time of crossing `T=b/v_(y)=(0.3xx5)/12xx1/10=1/8` hrs. Drift `x=v_(x)xxT=16/5xx1/8=0.4 km=400 m` So distance travelled by boat `S=sqrt(b^(2)+x^(2))=500 m` |
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| 1189. |
A swimmer crosses a flowing stream of width `omega` to and fro in time `t_(1)`. The time taken to cover the same distance up and down the stream is `t_(2)`. If `t_(3)` is the time the swimmer would take to swim a distance `2omega` in still water, then |
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Answer» Let (u) and (v) be the velocity of a swimmer in still warer and river water respectively. Then ltbRgt ` t_1 = 2 (d/(sqrt (u^2 -v^2))` ` t_2 = d/(u-v) + d/(u-v) = ( 2 ud)/ ( u^2- v^2)` and ` t_3 = (2 d)/u` :. T_2 t_3 = ( 2 ud)/(u^2 -v^2) xx ( 1d)/u = ( 4 d^2) /(u^2-v^2) = t_1^1`. |
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| 1190. |
A stick of length `L = 2.0 m` is leaned against a wall as shown. It is released from a position when `theta = 60^(@)`. The end A of the stick remains in contact with the wall and its other end B remains in contact with the floor as the stick slides down. Find the distance travelled by the centre of the stick by the time it hits the floor. |
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Answer» Correct Answer - `(pi)/(3)m` |
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| 1191. |
A body is in motion along a strainght line. As it crosses a fixed point a stop watch is started . The body travels a distance of ` 1.80 m` . In the first 3 second and 2.20 m in next 5 seconds. What will be the velocity at the end of 9 second ? |
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Answer» Didtance travelled in ( t+ 3) sec =1 . 8 , Distance traavelled in ( t+ 3+ 5) sec ` = 1.8 + 2.2 = 4.0 m`. |
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| 1192. |
If ` vec R = vec A- vec B` , and ` theta` is the smaller angle between ` vec A ` and ` vec B` , show that ltbRgt ` R^2 = A^2 + B^2 -2 AB cos theta`. |
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Answer» Given, ` vec R = vec A- vec B= vec A+ (- vec B), As ` theta` is the angle between ` vec A ` and ` vec B` , therefore, angle betwwen ` vec A` and ` vec -B` is ( 180^2-theta)`. Using prarallelogram law of vectrs addition ltbRgt or ` R^2 =A^2 + B^2 + 2 AB cos ( 180^2- theta)` ` = A^2 + B^2 -2 AB cos theta` . |
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| 1193. |
The value of tremperature caon be positive or negative. ltBrgt Temperature is a vector quantityA. (a) Statement-1 is true , Statement-2 is true , Statvement -2 is correct explanation of Statement-1 .B. (b) Statement-1 is true , statement -2 is true , statement -2 is not coerrecrt explanation of Statement-1.C. (C ) Statement-1 is true , Statement-2 is false.D. (d) Statement-1 is false , Statement-2 is true. |
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Answer» Correct Answer - A Here, the Assertion is true but temperature is a scalar quanitiy and not a vector , hence Teason is wrong. |
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| 1194. |
For what value of (a) are the vectors ltbRgt ` vec A= 2 a hat i-2 hat j + hat k` and ` vec B a hat I + a hat j -4 hat k` perpecdicular to each other ? |
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Answer» If ` vec A ` is perpendicular to `brv B`, then vec A, vec B =0` ltbRgt ( 2 a hat i-2 aht j + ahtk).(a hat i+ a hat j- 4 hat k) =0` ltbRgt or` `2 a^2 -2 a - 4 =0 ` or `2 a^2 -4 a+ 2 a- 4 =0` or ` 2 a (a- 2) + 2 9a -2) =0` ltbRgt pr ` (2 a + 2) ( a-2) =0` If ` a- 2 =0` then ` a= 2` If ` 2 a = 2 =0` theta ` a=- 1`. |
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| 1195. |
For what vlue of a, ` vec A = 2 hat i+ a hat j + hat k`, is perpendicular to ` vec B= 4 hat i-2 hat k`. |
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Answer» When vectors are perpendicular, their dot product is zero , i.e. ` vec A. vec B=0` or ` ( 2hat i+ a hat j= hat k). ( 4 aht i-2 hat j-2 aht k) =0` or ` 8 - 2 a - 2 =0 ` or ` a=3`. |
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| 1196. |
Two cities `A and B` are connected by a regular bus service with buses plying in either direction every `T` seconds. The speed of each bus is uniform and equal to `V_b`. A cyclist cycles from `A to B` with a uniform speed of `V_c`. A bus goes past the cyclist in `T_1` second in the direction `A to B` and every `T_2` second in the direction `B to A`. ThenA. `T_1 = (V_b T)/(V_b + V_c)`B. `T_2 = (V_b T)/(V_b - V_c)`C. `T_1 = (V_b T)/(V_b - V_c)`D. `T_2 = (V_b T)/(V_b + V_c)` |
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Answer» Correct Answer - C::D (c.,d.) Distance between two buses on road is `V_bT`. For `A to B` direction : Distance = Relative velocity xx Time `V_b T=(V_b - V_c)T_1 rArr T_1 = (V_b T)/(V_b - V_c)` For `B to A` direction : `V_b T =(V_b + V_c)T_2 rArr T_2 = (V_b T)/(V_b + V_c)`. |
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| 1197. |
Two cars are moving in the same direction with the same speed with separation 40 m.In order to overtake, the driver of the second car accelerates and he overtakes the first car in 5 seconds.What is the acceleration produced by the accelerator ? |
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Answer» Correct Answer - `3.2 ms^(-2)` |
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| 1198. |
A street car moves rectilinearly from station A to the next station B with an acceleration varying according to the law `f=a-bx`, where a and b are constants and x is the distance from station A. The distance between the two stations & the maximum velocity are :A. `x=(2a)/b, v_("max")=a/sqrt(b)`B. `x=b/(2a), v_("max")=a/b`C. `x=a/(2b), v_("max")=b/sqrt(a)`D. `x=a/b, v_("max")=sqrt(a)/b` |
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Answer» Correct Answer - A |
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| 1199. |
A graph is drawn between the displacement and time taken for the oscillations of a simple pendulum is 2s. The length of the penddulum is _______ m ( if g = 10 `s^(-2) and pi^(2) =10`)A. 2B. 1C. 0.5D. 0.25 |
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Answer» Correct Answer - B The frequency of the bob, i.e, the number of oscillations completed in one second = `1//2 Hz= 0.5 Hz` |
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| 1200. |
A graph is drawn between the displacement and time taken for the oscillations of a simple pendulum. The frequency of oscillation is ______ HZ.A. 0.5B. 2hC. 50D. 100 |
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Answer» Correct Answer - A Frequency of the bob, i.e, number of oscillation completed in one second = 1/2 Hz= 0.5 Hz |
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