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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.
| 1. |
A grasshopper can jump a maximum distance `1.6 m`. It spends negligible time on the ground. How far can it go in `10 s` ? |
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Answer» Correct Answer - 6 Maximum range `(theta = 45^@) = u^2/g = 0.8 u = 2sqrt2 m` Distance covered in `3 s = (u cos 45^@) (3)` `= (2sqrt(2))(1/(sqrt2))(3) = 6m`. |
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| 2. |
If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related asA. `y = 4h (t/T)(1-t/T)`B. `y = 4h (x/R)(1-x/R)`C. `y = 4h (T/t)(1-T/t)`D. `y = 4h (R/x)(1-R/x)` |
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Answer» Correct Answer - A::B a,b. `T = (2usintheta)/g , h = (u^2sin^2theta)/(2g), R = (u^2sin2theta)/g` `y = usinthetat - 1/2 "gt"^2` `=4ht[(usintheta)/(4h) - ("gt")/(8h)]` `= 4ht[ (usintheta)/(4(u^2sin^2theta//2g)) - ("gt")/(8(u^2sin^2theta//2g))]` `= 4ht[1/(2usintheta//g) - t/((2usintheta//g)^2)]` `4ht[1/T - t/T^2] = 4h (1/T)(1-t/T)` Now `x = ucos theta t, R = u cos theta T` `rArr t/T = x/R, so y = 4h(x/R)(1-x/R)`. |
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| 3. |
A motor boat is to reach at a point `30^@` upstream on the outer side of a river flowing with velocity `5 ms^(-1)`. The velocity of motor boat with respect to water is `5(sqrt3) ms^(-1)`. The driver should steer the boat at an angle. A. `30^@` w.r.t. the line of destination from the starting point.B. `60^@` w.r.t. normal to the bank.C. `120^@` w.r.t. stream directionD. None of these |
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Answer» Correct Answer - C c. The velocity of motor boat is given as `vecv_m = vecv_(mw) + vecv_(w)` `5/(sintheta) = (5sqrt(3))/(sin120^@) rArr sin theta = 1//2 rArr theta = 30^@`. |
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| 4. |
A ball is projected from the origin. The x- and y-coordinates of its displacement are given by `x = 3t and y = 4t - 5t^2`. Find the velocity of projection `("in" ms^(-1))`. |
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Answer» Correct Answer - 5 `v_x = (dx)/(dt) = 3 and v_y = (dy)/(dt) = 4 - 10t = 4 - 10(0) = 4` `v = [v_x^(2) + v_(y)^(2)]^(1//2) = [3^2 + 4^2]^(1//2) = 5ms^(-1)`. |
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| 5. |
The acceleration - velocity graph of a particle moving in a straight line is shown in figure. Then the slope of the velocity- displacement graph A. increases linearlyB. decreases linearlyC. is constantD. increases parabolically |
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Answer» Correct Answer - C c. Acceleration- velocity equation from the graph is a = kv, where k is the slope of given line. This can also be written as `v(dv)/(ds) = kv rArr (dv)/(ds) = k` i.e. slope of velocity - displacement graph is same as slope of acceleration- velocity graph which is constant. |
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| 6. |
Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be `0^@ltthetalt180^@`. During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions: A particle is projected from the origin in the x-y plane. The acceleration of particle in negative y-direction is `alpha`. If equation of path of the particle is `y = ax - bx^2`, then initial velocity of the particle isA. `sqrt(alpha/(2b))`B. `sqrt((alpha(1+a^2))/(2b))`C. `sqrt(alpha/(a^2))`D. `sqrt((alphab)/(a^2))` |
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Answer» Correct Answer - B b. `y = ax - bx^2` or `(dy)/(dt) = a(dx)/(dt) - 2bx(dx)/(dt) = (a-2bx) (dx)/(dt)` Initially `x=0,v_y = av_x` `(d^2y)/(dt^2) = (a-2bx) (d^2x)/(dt^2)+[-2b(dx)/(dt)]` `= (a-2bx)(d^2y)/(dt^2) - 2b ((dx)/(dt))^2` At `t = 0 , (d^2y)/(dt^2) = -alpha, x= 0, (d^2x)/(dt^2) =0` `rArr -alpha = -2b((dx)/(dt))^2 = -2b(v_x)^2 or v_x = sqrt(a/(2b))` `rArr v = sqrt(v_(x)^(2) + v_(y)^(2)) = v_x sqrt(1+a^2) = sqrt((alpha(1+a^2))/(2b))` . |
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| 7. |
A particle moves in a circle of radius R. In half the period of revolution its displacement is ………… and distance covered is ………. . |
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Answer» Correct Answer - `2pi,pir` Displacement = 2r. Distance `= pir`. |
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| 8. |
Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be `0^@ltthetalt180^@`. During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions: A body is projected at angle of `30^@` and `60^@` with the same velocity. Their horizontal ranges are `R_1` and `R_2` and maximum heights are `H_1 and H_2`, respectively, then A body is projected at angle of `30^@` and `60^@` with the same velocity. Their horizontal ranges are `R_1` and `R_2` and maximum heights are `H_1` and `H_2`, respectively, thenA. `R_1/R_2gt1`B. `H_1/H_2gt1`C. `R_1/R_2lt1`D. `H_1/H_2lt1` |
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Answer» Correct Answer - D d. Ranges will be same .`H_1/H_2 = (sin^2theta_1)/(sin^2theta_2)`. |
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| 9. |
Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be `0^@`lt`theta`lt`180^@`. During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions: An object is projected from origin in x-y plane in which velocity changes according to relation `vecv = a hati + bx hatj`. Path of particle isA. HyperbolicB. CircularC. EllipticalD. parabolic |
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Answer» Correct Answer - D d. `vecv = ahati + bxhatj , v_x = a or dx = adt rArr x = at` `v_y = bx = bat or y = (bat^2)/2` `rArr y = (ba)/2(x/a)^2 rArr y prop x^2 rarr parabolic` |
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| 10. |
Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take `g = 10 ms^(-2)`. Projection angle with the horizontal is :A. `tan^(-1) (4/5)`B. `tan^(-1) (2/3)`C. `tan^(-1) (5/4)`D. `tan^(-1) (1/2)` |
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Answer» Correct Answer - C Time of flight (T) = 1s `(2u_y)/g = 1s rArr u_y = 5ms^(-1)` Range (R) = 4m. `(2(u_x)(u_y)).g = 4 rArr u_x = 4 ms^(-1)` `u = sqrt(u_x^2 + u_(y)^(2)) = sqrt (41) ms^(-1)` `theta = tan^(-1) (u_y/u_x) = tan^(-1) (5/4)` `H = (u_y^2)/(2g) = (5)^2/20 = 5/4 = 1.25m`. |
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| 11. |
Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take `g = 10 ms^(-2)`. , The projection speed is :A. `sqrt(37)ms^(-1)`B. `sqrt(41)ms^(-1)`C. `sqrt(14)ms^(-1)`D. `sqrt(40)ms^(-1)` |
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Answer» Correct Answer - B Time of flight (T) = 1s `(2u_y)/g = 1s rArr u_y = 5ms^(-1)` Range (R) = 4m. `2(u_x)(u_y).g = 4 rArr u_x = 4 ms^(-1)` `u = sqrt(u_x^2 + u_(y)^(2)) = sqrt (41) ms^(-1)` `theta = tan^(-1) (u_y/u_x) = tan^(-1) (5/4)` `H = (u_y^2)/(2g) = (5)^2/20 = 5/4 = 1.25m`. |
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| 12. |
For a particle moving along x-axis, a scaled x-6 graph is shown in figure. Mark the correct statement (s). A. Speed of the particle is greatest at C.B. Speed of the particle is greatest at B.C. Particle is speeding up in region marked CD.D. Particle is speeding up in region marked AB. |
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Answer» Correct Answer - B::C::D b,c,d. Point of steepest slope corresponds to the maximum speed. Particle will speed up when the directions of acceleration and velocity are same. For region AB, both the acceleration and velocity are positive while for CD both are negative, so particle is speeding yp in these regions. |
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| 13. |
Mark the correct statement (s).A. A particle can have zero displacement and non-zero average velocity.B. A particle can have zero displacement and non-zero velocity.C. A particle can have zero acceleration and non-zero velocity.D. A particle can have zero velocity and non-zero acceleration. |
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Answer» Correct Answer - B::C::D b,c,d. Based on thought. |
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| 14. |
For a particle moving along the x-axis, mark the correct statement(s).A. if x is positive and is increasing with the time, then average velocity of the particle is positive.B. if x is negative and becoming positive after some time, then the velocity of the particle is always positive.C. If x is negative and becoming less negative as time passes, then the average velocity of the particle is positive.D. If x is positive and is increasing with time, then the velocity of the particle is always positive. |
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Answer» Correct Answer - A::C::D a,c,d. Self - explanatory. |
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| 15. |
For a particle moving along the x-axis, x-t graph is as given as graph in figure. Mar the correct statement (s). `(#BMS_V01_MAA_E01_052_Q01.png" width="80%">A. Initial velocity of the particle is zero.B. For BC, acceleration is positive and for DE, acceleration is negative.C. For EF, the acceleration is positive.D. Velocity becomes zero three times in the motion. |
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Answer» Correct Answer - A::C::D a,c,d. Velocity is given by the slope of the x-t graph. Here the slope is zero at A and at peak of region CD and bottom most point of EF. For AB and EF, acceleration is positive. For these regions, graph is concave up. For BC and DE, acceleration is zero as here velocity is constant. For CD, acceleration is negative as graph is concave down for this region. |
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| 16. |
A particle starts moving rectilinearly at time t = 0 such that its velocity v changes with time t according to the equation `v = t^2-t`, where t is in seconds and v is in `m s^(-1)`. The time interval for which the particle retrads (i.e., magnitude of velocity decreases)isA. `t lt 1//2`B. `1//2 ltt lt1`C. `tgt1`D. `tlt1//2 and tgt1` |
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Answer» Correct Answer - B b. ` v= t^2-t rArr a = 9dv)/(dt) = 2t - 1 ` For retardation, `avlt0` `rArr (t^2-t)(2t-1)lt0` ` rArr t(t-1)(2t-1)lt0` This is possible for `1/2lttlt1`. |
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| 17. |
If acceleration is constant and initial velocity of the body is 0, then choose the correct statement. Symbols have their usual meaning.A. `v prop sqrt(t)`B. `v prop sqrt(x)`C. `v prop t`D. `v prop x^2` |
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Answer» Correct Answer - B::C b,c. `u = 0, a= a_0` `v_2 - 0_2 = 2a_0x` `rArr v = sqrt(2a_0x) rArr v prop sqrt(x)` `v = u + at rArr v prop t ` . |
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| 18. |
A particle moves in a straight line. Its position ( in m) as function of time is given by `x = (at^2 + b)` What is the average velocity in time interval ` t = 3s to t = 5s in ms^(-1)`. (where a and b are constants and a `= 1ms^(-2), b = 1m`). |
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Answer» Correct Answer - 8 `V_(av) = (x_(f)- x_(i))/(t_(f)-t_(i)) = ((1 xx 5^2 +1)-(1xx3^2+1))/(5-3) = 16/2 = 8 ms^(-1)`. |
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| 19. |
A particle can move only along x-axis. Three pairs of initial and final positions of particle at two successive times are given Pair Initial positions Final position 1 (-3)m (+5)m 2 (-3) m (-7)m 3 (+7) (-3), Find the sum of magnitudes of displacement in the pairs which give negative displacement in m. |
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Answer» Correct Answer - 14 `s_1 = 5(-3) = 8m` `s_2 = -7-(-3) = -4m` `s_3 = -3-7 = -10m` Required value = `(s_2+s_3) = (-14) = 14`. |
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| 20. |
Figure shows the velocity-displacement curve for an object moving along a straight line. At which of the points marked is the object speeding up? A. 1B. 2C. 1 and 3D. 1,2, and 3 |
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Answer» Correct Answer - A a. From a = v `dv//ds`, we can find the sign of acceleration at various points. V is positive for all three points 1,2,and 3. `dv//ds` is positive for point 1, zero for point 2, negative for point 3. So, only for point 1, velocity and acceleration have same sign, so the object is speeding up at point 1 only. |
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| 21. |
From the top of a tower of height 200 m, a ball A is projected up with speed `10ms^(-1)` and 2 s later, another ball B is projected vertically down with the same speed. ThenA. Both A and B will reach the ground simultaneouslyB. Ball A with hit the ground 2 s later than B hitting the groundC. Both the balls will hit the ground with the same velocityD. Both will rebound to the same height from the ground, if both have same coefficient of restitution. |
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Answer» Correct Answer - A::C::D a,c,d. Total displacement ball A in 2s, `S = ut - 1/2"gt"^2 = 10 xx 2 - 1/2 xx 10 xx (2)^2 = 0` Hence, both A and B will reach the ground simultaneously and strike with the same velocity. |
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| 22. |
Two particles A and B are placed as shown in figure. The particle A, on the top of tower, is projected horizontally with a velocity u and particle B is projected along the surface towards the tower, simultaneously. If both particles meet each other, then the speed of projection of particles B is [ignore any friction] A. `dsqrt(g/(2H)) - u`B. `d sqrt(g/(2H))`C. `dsqrt(g/(2H))+u`D. u |
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Answer» Correct Answer - A a. Time taken by particles to collide `t = sqrt((2H)/(g))` Then `u sqrt((2H)/(g))+Vsqrt((2H)/g) = d` `rArr u + V = d sqrt(g/(2H)) or v = d sqrt(g/(2H)) - u `. |
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| 23. |
Two particles are thrown horizontally in opposite directions with velocities u and 2u from the top of a high tower. The time after which their radius of curvature will be mutually perpendicular isA. `(sqrt2) u/g`B. `2 u/g`C. `1/(sqrt2) u/g`D. `1/2 u/g` |
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Answer» Correct Answer - A a. The radius of curvatures will be mutually perpendicular only when the velocity vectors will be mutually perpendicular, i.e., after the time `t = sqrt(2) u/g.` |
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| 24. |
A ball is projected from ground with speed u, at an angle `theta` above horizontal. Let v be its speed at any moment t and s be the total distance covered by it till this moment, the correct graph (s) `is//are`.A. B. C. D. |
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Answer» Correct Answer - C::D c,d. Speed first decreases. Then increases. Distance covered increases continuously. |
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| 25. |
Two particles are thrown simultaneously from points A and B with velocities `u_1 = 2ms^(-1) and u_2 - 14 ms^(-1)`, respectively, as shown in figure. The direction (angle) with horizontal at which B will appear to move as seen from A isA. `37^@`B. `53^@`C. `15^@`D. `90^@` |
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Answer» Correct Answer - B b. `tanphi = (6sqrt2)/(8sqrt2) rArr tanphi = 3/4 rArr phi = 37^@` . |
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| 26. |
Two particles are thrown simultaneously from points A and B with velocities `u_1 = 2ms^(-1) and u_2 - 14 ms^(-1)`, respectively, as shown in figure. The relative velocity of B as seen from A inA. `-8sqrt(2)hati + 6(sqrt2)hatj`B. `4sqrt(2)hati + 3(sqrt3)hatj`C. `3sqrt(5)hati + 2(sqrt3)hatj`D. `3sqrt(2)hati + 4(sqrt3)hatj` |
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Answer» Correct Answer - A a. `V_(B,A) = V_B-V_A` `= -14 cos 45^@hati + 14sin45^@hatj - (2cos45^@hati)-2sin45^@hatj` `= -16/(sqrt2)hati + 12/(sqrt2)hatj = -8sqrt(2)hati + 6sqrt(2)hatj`. |
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| 27. |
Two particles P and Q are projected simultaneously away from each other from a point A as shown in figure. The velocity of P relative to Q in `ms^(-1)` at the instant when the motion of P is horizontal is A. `10sqrt(4-sqrt3)`B. `20sqrt(4-sqrt3)`C. `10sqrt(4+sqrt3)`D. `20sqrt(4+sqrt3)` |
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Answer» Correct Answer - B b. Intially: `vecu_(P) = 20cos60hati + 20sin60^@hatj = 10hati + 10(sqrt3)hatj` `vecu = 20(sqrt2)[cos45^@(-hati)+sin45^@hatj] = -20hati + 20hatj` Initial relative velocity: `vec(u_(p//Q)) = vecu_(p)-vecu_(Q) = 30hati+(10(sqrt3)-20)hatj` `u_(p//Q) = sqrt(30^2+(10(sqrt3)-20)^2) = 20 sqrt(4-sqrt(3)) ms^(-1)` This relative velocity will remain same till both the particles are in air, because relative acceleration is zero. |
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| 28. |
Two bodies 1 and 2 are projected simultaneously with velocities `v_1 = 2ms^(-1) and v_2 = 4ms^(-1)` respectively. The body 1 is projected vertically up from the top of a cliff of height h = 10 and the body 2 is projected vertically up from the bottom of the cliff. If the bodies meet, find the time (in s) of meeting of the bodies. |
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Answer» Correct Answer - 5 Let the particle meet after a time t. First of all, we choose the point of collision above the top of the cliff. For the first particle, `s = s_1, v_0 = v_1, a = -g` Then, `s_1 = v_1t - 1/2 "gt"^2` .........(i) For the second particle, `s = s_2, v_0 = v_1, a= -g` Then , `s_2 = v_2t - 1/2 "gt"^2` ..........(ii) Referring to the figure, `s_2 - s_1 = h`......(iii) Substtituting `s_1` from Eq. (i), s_2 from Eq. (ii) in Eq. (iii), we have ` t= h/(v_2-v_1) = 10/(4-2) = 5s`. |
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| 29. |
Shots are fired simultaneously from the top and bottom of a vertical cliff with the elevation `alpha = 30^@`, beta `= 60^@`, respectively. The shots strike an object simultaneously at the same point. If `a = 10 (sqrt3)` m is the horizontal distance of the object from the cliff, then the height h of the cliff is A. 30 mB. 45 mC. 60 mD. 90 m |
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Answer» Correct Answer - C c. `u_1cos alpha = u_2cosbeta` `30sqrt(3) = (u_(2)^(2)sin2beta)/g rArr 30sqrt(3) = (u_(2)^(2)sqrt(3))/(2g)` `rArr u_(2)^(2) = 600 or u_2 = 10sqrt(6)` `t_2 = (2u_2sin60^@)/g = (2(10sqrt(6))sqrt(3)//2)/10 rArr t_2 = sqrt(18)` `u_1 = (10sqrt(6)cos60^@)/(cos30^@) = 10sqrt2` `-h = 10sqrt(2)sin 30^@sqrt(18) - 1/2 10 (sqrt18)^2` `=30-90 rArr h = 60m` . |
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| 30. |
A shot is fired at an angle `theta` to the horizontal such that it strikes the hill while moving horizontally. Find the initial angle of projection `theta`. A. `tan theta = 2/5`B. `tan theta = 3/8`C. `tan theta = 3/2`D. None of these |
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Answer» Correct Answer - C c. `(usintheta)/g = (2usin(theta-37^@))/(gcos37^@)` `sin theta = (2(sintheta)(4/5)-2costheta(3/5))/(4/5)` `4sintheta = 8 sin theta - 6 cos theta` `4 sin theta = 6 cos theta rArr tan theta = 3/2 ` . |
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| 31. |
Consider a shell that has a muzzle velocity of `45ms^(-1)` fired from the tail gun of an airplane moving horizontally with a velocity of `215 ms^(-1)` . The tail gun can be directed at any angle with the vertical in the plane of motion of the airplane. The shell is fired when the plane is above point A on ground, and the plane is above point B on ground when the shell hits the ground. (Assume for simplicity that the Earth is flat)A. Shell may hit the ground at point A.B. Shell may hit the ground at point B.C. Shell may hit a point on earth which is behind point A.D. Shell may hit a point on earth which is behind point B. |
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Answer» Correct Answer - B::D b,d. `vecV_(s//G) = vevV_(s//A) + vecV_(A//G)` Horizontal component of velocity of shell will be in the direction of velocity of air plane. |
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| 32. |
A particle moves along positive branch of the curve, `y = x/2, where x = t^3/3`, x and y are measured in meters and t in seconds, thenA. The velocity of particle at `t = 1 s is hati + 1/2 hatj`.B. The velocity of particle at `t = 1 s is 1/2 hati + hatj`.C. The acceleration of particle at ` t= 2 s is 2hati+hatj`.D. The acceleration of particle at ` t= 2 s is hati+ 2hatj`. |
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Answer» Correct Answer - A::C a,c. `(dx)/(dt) = t^2` ……………..(i) `y = 1/2 t^3/3 rArr (dy)/(dt) = t^2/2`…………..(ii) `t = 1, v_x = 1, v_y = 1/2 rArr vecv = hati + 1/2 hatj`. `(d^2x)/(dt^2)= 2t` ………(iii) `(d^2y)/(dt^2) = t` ……(iv) At t = 1 s, `a_x = 2 and a_y = 1` `rArr veca = 2hati + hatj`. |
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| 33. |
Two balls are projected from points A and B in vertical plane as shown in figure. AB is a straight vertical line. The balls can collide in mid air if `v_1//v_2` is equal to A. `(sintheta_1)/(sintheta_2)`B. `(sintheta_2)/(sintheta_1)`C. `(costheta_1)/(costheta_2)`D. `(costheta_2)/(costheta_1)` |
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Answer» Correct Answer - D d. If the particles collide in mid air, they travel same displacement in horizontal direction. So their velocity components along horizontal should be same, i.e., `v_1costheta_1 = v_2costheta_2`. |
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| 34. |
A particle is moving along the x-axis whose acceleration is given by ` a= 3x-4`, where x is the location of the particle. At t = 0, the particle is at rest at `x = 4//3m`. The distance travelled by the particles in 5 s isA. zeroB. 42 mC. InfiniteD. None of these |
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Answer» Correct Answer - A a. At `t = 0 , v=0, x = 4//3m` `a = (4xx3)/3 - 4 = 0` As both the particle velocity and acceleration are zero at t = 0, it will always remain at rest and hence distance travelled at any time interval would be zero. |
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| 35. |
Two boys P and Q are playing on a river bank. P plans to swim across the river directly and come back. Q plans to swim downstream by a length equal to the width of the river and then come back. Both of them bet each other, claiming that the boy succeeding in less time will win. Assuming the swimming rate of both P and Q to the same, it can be concluded thatA. P winsB. Q winsC. A draw takes placeD. Nothing certain can be stated. |
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Answer» Correct Answer - A a. Let d be the river width and u and v the speeds of water current. for P, time taken `t_1 = (2d)/(sqrt(v^2-u^2))` ……….(i) For `Q`, time taken, `t_2 = d[1/(v+u) + 1/(v-u)]………(ii)` Dividing Eq. (i) by (ii), we get `t_1/t_2 = sqrt (1-(u/v)^2)lt1 rArr t_1ltt_2` . |
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| 36. |
Two guns on a battleship simultaneously fire two shells with same speed at enemy ships. If the shells follow the parabolic trajectories as shown in figure. Which ship will get hit first? A. AB. BC. both at same timeD. need more information |
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Answer» Correct Answer - B b. Time of flight depends upon maximum height and maximum height for A is large in comparison to B. |
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| 37. |
The velocity - time graph of two bodies A and B is shown in figure. Choose correct statement. A. acceleration of Bgt acceleration of AB. acceleration of A gt acceleration of BC. both are starting from same pointD. A covers greater distance than B in the same time. |
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Answer» Correct Answer - B::D b,d. Slope of A is greater than slope of B. `:. a_Agta_B` `S = 1/2 at^2 (:. u= 0)` `:. S_(A)gtS_(B)`. |
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| 38. |
A particle is moving in a circular path of radius 1 m. under the action of a centripetal force, the speed `sqrt2pi ms^(-1)` of the particle is constant. Find the average of the velocity `("in" ms^(-1))` between A and B. |
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Answer» Correct Answer - 4 `v_(av) = (rsqrt2)/(T//4) = (rsqrt2xx4v)/(2pir) = (2sqrt2v)/pi = (2sqrt(2)(sqrt2)pi)/pi = 4ms^(-1)`. |
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| 39. |
Acceleration of particle moving rectilinearly is `a = 4-2x`(where x is position in metre and a in ms^(-2))`. It si at instantaneous rest at x = 0. At what position `x 9`in meter) will the particle again come to instantaneous rest? |
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Answer» Correct Answer - 4 `(vdv)/(dx) = 4-2x` `int_(0)^(v) vdv = int_(0)^(x) (4-2x)dx rArr v^2/2 = 4x - x^2` when `v = 0 , 4x - x^2 = 0` `x = 0, 4` `rArr Thus, at x = 4, the particle will again come to rest. |
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| 40. |
Figure shows path followed by a particle and position of a particle at any instant. Four different students have represented the velocity vectors and acceleration vectors at the given instant. Which vector diagram can not be true in any situation? (In each figure velocity is tangential to the trajectory). A. SitaB. GitaC. RamD. Shyam |
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Answer» Correct Answer - D d. In curved path is always within the curvature of trajectory. |
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| 41. |
The x-t graph of a particle moving along a straight line is shown in figure The speed-time graph of the particle is correctly shown byA. B. C. D. |
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Answer» Correct Answer - C c. Speed is always positive. So we can obtain speed - time graph from velocity-time graph by rotating the part of velocity- time graph below time axis by `180^@`. |
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| 42. |
Ram Shyam are walking on two perpendicular tracks with speed `3ms^(-1)` and `4 ms^(-1)`, respectively. At a certain moment `(say t = 0s)`, Ram and Shyam are at 20 m and 40 m away from the intersection of tracks, respectively, and moving towards the intersection of the tracks. During the motion the magnitude of velocity of ram with respect to Shyam isA. `1 ms^(-1)`B. `4 ms^(-1)`C. `5 ms^(-1)`D. `7 ms^(-1)` |
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Answer» Correct Answer - C c. `vecv_(R,S) = vecv_R - vecv_S - vecv_R + (-vecv_S)` `|vecv_(R,S)| = sqrt(3^2 + 4^2) = 5ms^(-1)` `tan theta |vecv_R|/|vecv_S| = 3/4` Hence, `v = 37%@` |
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| 43. |
Ram Shyam are walking on two perpendicular tracks with speed `3ms^(-1) and 4 ms^(-1)`, respectively. At a certain moment (say t = 0s), Ram and Shyam are at 20 m and 40 m away from the intersection of tracks, respectively, and moving towards the intersection of the tracks. The time t when they are at shortest distance from each other subsequently is -A. 8.8 sB. 12 sC. 15 sD. 44 s |
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Answer» Correct Answer - A a. `tan 37^@ = 8/(BC)` `3/4 - 8/(BC)` `BC = 32/3m` `sin 53^@ = 80/(3AB), 4/5 = 80/(3xxAB)` `AB = (100)/3m` ` t = ((32/3 + 100/3))/5 = 132/15 = 8.8s`. |
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