InterviewSolution
Saved Bookmarks
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.
| 601. |
A gun is firing bullets with velocity `v_0` by rotating it through `360^@` in the horizontal plane. The maximum area covered by the bullets is |
|
Answer» The maximum area will be equal to the area of the circle with radius equal to the maximum range of projectile. Maximum area, `pi r^2 = pi (R_(max))^2 = pi ((u^2)/(g^2)) = pi (u^4)/(g^2)` `[As r = R_(max) = u^2//g for theta = 45^@]`. |
|
| 602. |
A particle is thrown with velocity `u` at an angle `prop` from the horizontal. Another particle is thrown with the same velocity at an angle `prop` from the verticle. What will be the ratio of times of flight of two particles ? |
|
Answer» For first particles angle of projection from the hortizontal is `prop`. So `T_1 = (2usinprop)/(g)` For second particle, angle of projection from the vertical is `prop`. it mean from the horizontal is `(theta - prop)` `therefore T_(2) = (2 u sin(90 - prop))/(g) = (2 u cos prop)/(g)` So, ratio of time of flight `(T_1)/(T_2) = tan prop`. |
|
| 603. |
Two particles are located on a horizontal plane at a distance `60 m`. At `t = 0` both the particles are simultaneously projected at angle `45^@` with velocities `2 ms^-1 and 14 m s^-1`, respectively. Find (a) Minimum separation between them during motion. (b) At what time is the separation between them minimum ? . |
|
Answer» Correct Answer - [6 m] |
|
| 604. |
Two particles `A and B` are moving with constant velocities `v_1 and v_2`. At `t = 0`, `v_1` makes an angle `theta_0` with the line joining `A and B` and `v_2` makes an angle `theta_2` with the line joining `A and B`. Find their velocity of approach. . |
|
Answer» Velocity of approach is the relative velocity along line `AB` `v_(A P P) = v_1 cos theta_1 + v_2 cos theta_2`. |
|
| 605. |
The height `y` nad the distance `x` along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t - 5t^2) m` and `x = 6t m`, where `t` is in seconds. The velocity with which the projectile is projected at `t = 0` is.A. `8 m s^-1`B. `6 m s^-1`C. `10 m s^-1`D. Not obtainable from the data. |
|
Answer» Correct Answer - C ( c) `x = 6t, u_x = (dx)/(dt) = 6` `y = 8t - 5t^2, v_y = (dy)/(dt) = 8 - 10 t` `v_y_(t = 0) = 8 ms^-1` `u = sqrt(u_x^2 + u_y^2) =sqrt(6^2 + 8^2) = 10 ms^-1`. |
|
| 606. |
A body is thrown horizontally with a velocity (v) from a towr ` H` metre high .After how much time and at what distance from the base of the towrt will the body strike the ground ? |
|
Answer» Time, ` t= sqrt 2 H//g` , Horizontal distance, ` x =ut =v sqrt2 H//g`. |
|
| 607. |
Statement-I : In a free fall, the initial velocity of a body may or may not be zero. Statement-II : A heavy body falls at a faster rate as compared to a light body.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
|
Answer» Correct Answer - C Free fall implies that the particle moves only in presence of gravity. |
|
| 608. |
Distance is a scalar quantity. Displacement is a vector quqntity. The magnitude of displacement is always less than or equal to distance. For a moving body displacement can be zero but distance cannot be zero. Same concept is applicable regarding velocity and speed. Acceleration is the rate of change of velocity. If acceleration is constant, then equations of kinematics are applicable for one dimensional motion under the gravity in which air resistance is considered, then the value of acceleration depends on the density of medium. Each motion is measured with respect of frame of reference. Relative velocity may be greater`//` smaller to the individual velocities. A particle moves from A to B. Then the ratio of distance to displacement is :- A. `pi/2`B. `2/pi`C. `pi/4`D. `1:1` |
|
Answer» Correct Answer - A `("Distance")/("Dispacement")=(pid//2)/d=pi/2` |
|
| 609. |
Distance is a scalar quantity. Displacement is a vector quqntity. The magnitude of displacement is always less than or equal to distance. For a moving body displacement can be zero but distance cannot be zero. Same concept is applicable regarding velocity and speed. Acceleration is the rate of change of velocity. If acceleration is constant, then equations of kinematics are applicable for one dimensional motion under the gravity in which air resistance is considered, then the value of acceleration depends on the density of medium. Each motion is measured with respect of frame of reference. Relative velocity may be greater`//` smaller to the individual velocities. A particle is moving along the path `y=4x^(2)`. The distance and displacement from `x=1` to `x=2` is (nearly) :-A. `sqrt(150), 12`B. `sqrt(160), 20`C. `sqrt(200), 30`D. `sqrt(150), 20` |
|
Answer» Correct Answer - A `x_(1)=1, y_(1)=4, x_(2)=2, y_(2)=16` `:.` Displacement `=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))` `=sqrt(1^(2)+12^(2))=sqrt(145) equiv12m` |
|
| 610. |
Statement-I : When the direction of motion of a particle moving in a circular path is reversed the direction of radial acceleration still remains the same (at the given point). Statement-II : Particle revolves on circular path in any direction such as clockwise or anticlockwise the direction of radius accelerationis always towards the centre of the circular path.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
| Answer» Correct Answer - A | |
| 611. |
Statement-I : A cyclist must adopt a zig-zag path while ascending a steep hill. Statement-II : The zig-zag path prevent the cyclist to slip down.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
| Answer» Correct Answer - A | |
| 612. |
A shot is fired from a point at a distance of `200 m` from the foot of a tower `100 m` high so that it just passes over it horizontally. The direction of shot with horizontal is.A. `30^@`B. `45^@`C. `60^@`D. `70^@` |
|
Answer» Correct Answer - B (b) `H = 100 m, R = 2 xx 200 = 400 m` `tan theta = (4 H)/(R ) rArr tan theta = (4 xx 100)/(400) = 1` `rArr theta = 45^@ [because (H)/( R) = (tan theta)/(4)]`. |
|
| 613. |
Identify the correct graph represeriting the motion of a particle along a straight line With constant acceleration with zero initial velocity.A. B. C. D. |
|
Answer» Correct Answer - A::D `v=at and x=1/2 at^2(u=0)`i.e. v-t graph is a straight line passing through origin and x-t graph a parabola passing through origin. |
|
| 614. |
In the above problem, what is the angle of projection with horizontal ?A. `tan^-1 (1//4)`B. `tan^-1 (4//3)`C. `tan^-1 (3//4)`D. `tan^-1 (1//2)` |
|
Answer» Correct Answer - B (b) `tan theta = (v_y)/(v_x) = (8)/(6) = (4)/(3) or theta = tan^-1 (4)/(3)`. |
|
| 615. |
The height `y` and the distance `x` along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t - 5t^2) m` and `x = 6tm`, where `t` is in seconds. The velocity with which the projectile is projected at `t = 0` is.A. `6 m s^-1`B. `8 m s^-1`C. `10 m s^-1`D. `14 m s^-1` |
|
Answer» Correct Answer - C ( c) `v_x = (dx)/(dt) = 6 and v_y = (dy)/(dt) = 8 - 10 t` Putting `t = 0` (Since we have to find initial velocity) `v_y = 8 - 10 xx 0 = 8` `v = sqrt(v_x^2 + v_y^2)) = sqrt(6^2 + 8^2) =10 ms^-1`. |
|
| 616. |
The figure shows the velocity (v) of a particle plotted against time (t). A. The particle changes its direction of motion at some point.B. The acceleration of the particle remains constantC. The displacement of the particle is zeroD. The initial and final speeds of the particle are the same |
|
Answer» Correct Answer - A::B::C::D For `tltT, v=-ve` For `tgtT, v=+ve` At `t=T,v=0` `:.` Particle change direction of velocity at `t=T` `S`=Net area of v-t garph=0 `a`=Slope of v-t graph =constant |
|
| 617. |
The figure shows the velocity (v) of a particle plotted against time (t). A. The particle changes its direction of motion at some point.B. The displacement of the particle remains constant.C. The displacement of the particle is zero.D. The initial and dinal speeds of the particle are the same. |
|
Answer» Correct Answer - A::B::C::D Particle changes direction of motion at `t=T`. Acceleration remains constant, becoause the velocity-time graph is a straigh line .Dislacement is zero, because net area is zero . Initial and final speeds are equal. |
|
| 618. |
The figure shows the velocity (v) of a particle plotted against time (t). A. The particle changes its direction of motion at some pointB. The acceleration of the particle remains constantC. The displacement of the particle is zeroD. The initial and final spedds of the particle are the same |
| Answer» Correct Answer - A::B::C::D | |
| 619. |
Aparticle moves with a speed `v` changes direction by an angle `theta`, without change in speed.A. the change in the magnitude in velocity is zeroB. the change in the magnitude fo its velocity is `2v sin theta//2`C. the magnitude of change in velcoity is `2vsin theta//2`D. the magnitude of change in velocity is `v(1-cos theta)` |
| Answer» Correct Answer - A::C | |
| 620. |
Let y =f(x) is a function. Its maxima (or) minima can be obtained by –(a) y = 0 (b) f(x) = 0(c) \(\frac{dy}{dx} = 0\)(d) \(\frac{d^2y}{dx^2} = 0\) |
|
Answer» Correct answer is (c) \(\frac{dy}{dx} = 0\) |
|
| 621. |
A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is- (a) 9 N (b) 16 N (c) 5 N (d) 7 N |
|
Answer» Correct answer is (c) 5 N |
|
| 622. |
A constant force is acting on a particle and also acting perpendicular to the velocity of the particle. The particle describes the motion in a plane. Then –(a) angular displacement is zero (b) its velocity is zero (c) it velocity is constant (d) it moves in a circular path |
|
Answer» (d) it moves in a circular path |
|
| 623. |
The range of the projectile depends – (a) The angle of projection (b) Velocity of projection (c) g (d) all the above |
|
Answer» (d) all the above |
|
| 624. |
The V – t graphs of two objects make angle 30° and 60° with the time axis. Find the ratio of their accelerations. |
|
Answer» \(\frac{a_1}{a_2}\) = \(\frac{tan 30}{tan 60}\) = \(\frac{1-\sqrt 3}{\sqrt 3}\) = \(\frac{1}{3}\) = 1:3 |
|
| 625. |
The following graphs represent velocity – time graph. Identity what kind of motion a particle undergoes in each graph |
|
Answer» (a) At all the points, slope of the graph is constant. ∴ \(\vec a\) = constant (b) No change in magnitude of velocity with respect to time ∴ \(\vec V\) = constant (c) Slope of this graph is greater than graph (a) but constant ∴ \(\vec a\) = constant but greater than the graph (a) (d) At each point slope of the curve increases. ∴ \(\vec a\) is a variable and object is in accelerated motion. |
|
| 626. |
Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects are denoted as R30° and R60° – Choose the correct relation from the following:(a) R30° = R60°(b) R30° = 4R60°(c) R30° = \(\frac{R60°}{2}\)(d) R30° = 2R60° |
|
Answer» Correct answer is (a) R30° = R60° |
|
| 627. |
The following table gives the range of a particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity, (g value).PlanetRangeJupiter50 mEarth75 mMars90 mMercury95 m |
|
Answer» Range = \(\frac{v^2}{g}\) sin 2θ ∴ g α \(\frac{1}{range}\) Ascending order of the planet with respect to their “g” is Mercury, Mars, Earth, Jupiter. |
|
| 628. |
A water fountain on the ground sprinkles water all around it . If the speed of water coming out of the fountain is ` v` , the total area around the fountain that gets wet is :A. `pi/2 v^(4)/g^(2)`B. `pi v^(2)/g^(2)`C. `pi v^(2)/g`D. `pi v^(4)/g^(2)` |
|
Answer» Correct Answer - 4 `R_("max")=u^(2)/g`, Area `=pir^(2)=(piu^(2)R_("max")^(2))/g^(2)` |
|
| 629. |
Read each staremnt below crefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre. (b) The velocity vector of a particle at a point is always aling the tagent to the path of the partile at that point. ltbRgt (c ) The acceleration vector of a particle in uniform circular motion acraged over one cycle is a null vector. |
|
Answer» (a) False. The net acceleration of a particle is towards the centre only in case of a uniform circulat motion. (b) True, because while veaving the circular path, the particle moves tagentially to the circualr path. (C ) True. The direction of accelration vector in a uniform circular motion is directed towards the centre of circular path. It is constantly changing with time. The resulatnt of all there vectors will be a zero vector. |
|
| 630. |
The location of a particale is changed. What can we say about the displacement and distance coverd by the particle?A. Both cannot be zeroB. One of the two may be zeroC. Both must be zeroD. Both must be equal |
|
Answer» Correct Answer - A If the location of a particle changes, then both distance and displacement must have some value. |
|
| 631. |
Read each staremnt below carefully and state with reasons and expamples if it is true or false , (a) with zero speed at an instant may have non-zero accelration at that instant (b) with zero speed may have non-zero velocity (c ) with positive speed must have zero accleration (d) with positive value of acceleration must be speeding up. |
|
Answer» (a) Ture, when a body is thrown vertically upwards inn the space, then at the highest point, the body has zero speed but has downward acceleration equal to the accelration due to gravity. (b) false, becomuse velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero. (c ) True, when a particle is moving along a straight line with a constant speed, its velocity remains constant with time . Therefore, acceleration (= change in velocity //time ) is zero. ltbRgt (d) The statement depends upon the chice of the instant of time taken as origin as origin, when the bdy is moving along a straight line with positive accelration. The velocity of the body at an instant of time (t) is ` v=u +at` The given statement is not correct if (a) is positive and (u) is negative, at the instant of time taken as origin. Then for all the times before the time for which (v) venishes, there is slowing down of the particle i.e. the speed of the particle keeps on decteasing with time. It happens when body is projected vertically upwards. However the given statement is true if u) is positive and (a) is positive, at the instant of time taken as origin. It is so when the body is falling verticall downwards. |
|
| 632. |
The magintude of displacemnt is equal to the distance coverd in a given interval of time if the particle .A. Moves with constant acceleration along any pathB. Moves with constant speedC. Moves in same direction with constant velocity or with variable velocity.D. Moves with constant velocity |
|
Answer» Correct Answer - C To have distance equal to magnitude of displacement. The particle has to move in the same derection. The velocity may or may not constant. |
|
| 633. |
The distance traversed by a moving particle at any instant is half of the product of its velocity and the time of travers. Show that the acceleration of particle is constant. |
|
Answer» Let at an instant ` t, v` be the velocity of the moving praticle and (S) be the distance travelled by the particle . As pre question. ` S =v t//2 ` …(i) Defferntiating it w.r.t. time (t), we have `(dS)/(dt) = 1/2 (d v)/(dt) xx t + v/2 ` or ` v =1/2 a xx t + v/2` (:. (dv)/(dt) =acceleration =a) or ` at =v` Differntiating it again w..r.t. time (t), we have ` (da)/(dt) xx t + a = (d v)/(dt) =a or (da)/(dt) xx t=0` or ` (da)/(dt) =0`. |
|
| 634. |
A `0.098-kg` block slides down a frictionless track as shown in (Fig. 5.208). . The horizontal distance `x` travelled by the block in moving from `A` to `C` is.A. `(1 + sqrt(3)) m`B. `(1 - sqrt(3)) m`C. `(sqrt(3) + 3)) m`D. g meter |
|
Answer» Correct Answer - C (c) `(1)/(2) mv^2 = mg(3 - 1) = 2 mg` [from conservation of energy] or `v = sqrt(4 g) = 2 sqrt(g)` Vertical component at A is `2 sqrt(g) sin 30^@ = sqrt(g)` Time of flight, `T = (2 v sin theta)/(g) = (2 sqrt(g))/(g) = (2)/(sqrt(g))` Using `S = ut + (1)/(2) at^2`, we get `-1 = sqrt(g) t - (1)/(2) "gt"^2 or (1)/(2) "gt"^2 - sqrt(g) t - 1 = 0` `t = (sqrt(g) +- sqrt(g + 4 xx (1)/(2) g))/(2 xx (g)/(2)) t = (1 +- sqrt(3))/(sqrt(g))` Neglecting negative time, `t = (1 +- sqrt(3))/(sqrt(g))` `x = 2 sqrt(g) cos 30^@ [(1 +- sqrt(3))/(sqrt(g))] or x = (sqrt(3) + 3) m`. |
|
| 635. |
A `0.098-kg` block slides down a frictionless track as shown in (Fig. ). . The vertical component of the velocity of block at `A` isA. `sqrt(g)`B. `2 sqrt(g)`C. `3 sqrt(g)`D. `4 sqrt(g)` |
|
Answer» Correct Answer - A (a) `(1)/(2) mv^2 = mg(3 - 1) = 2 mg` [from conservation of energy] or `v = sqrt(4 g) = 2 sqrt(g)` Vertical component at A is `2 sqrt(g) sin 30^@ = sqrt(g)` Time of flight, `T = (2 v sin theta)/(g) = (2 sqrt(g))/(g) = (2)/(sqrt(g))` Using `S = ut + (1)/(2) at^2`, we get `-1 = sqrt(g) t - (1)/(2) "gt"^2 or (1)/(2) "gt"^2 - sqrt(g) t - 1 = 0` `t = (sqrt(g) +- sqrt(g + 4 xx (1)/(2) g))/(2 xx (g)/(2)) t = (1 +- sqrt(3))/(sqrt(g))` Neglecting negative time, `t = (1 +- sqrt(3))/(sqrt(g))` `x = 2 sqrt(g) cos 30^@ [(1 +- sqrt(3))/(sqrt(g))] or x = (sqrt(3) + 3) m`. |
|
| 636. |
A student performs an experiment to determine how the range of a ball depends on the velocity with which it is projected. The "range" is the distance between the points where the ball lends and from where it was projected, assuming it lands at the same height from which it was projected. It each trial, the student uses the same baseball, and launches it at the same angle. Table shows the experimental results. `|{:("Trail","Launch speed" (m//s),"Range"(m)),(1,10,8),(2,20,31.8),(3,30,70.7),(4,40,122.5):}|` Based on this data, the student then hypothesizes that the range, R, depends on the initial speed `v_(0)` according to the following equation : `R=Cv_(0)^(n)`, where C is a constant and n is another constant. Based on this data, the best guess for the value of n is :-A. `1/2`B. `1`C. `2`D. `3` |
|
Answer» Correct Answer - C `R=Cv_(0)^(n)` Putting data from table: `8=Cxx10^(n)` `rArr 31.8=Cxx20^(n)rArr 31.8/8=3.9 cong 4=2^(n) rArr n=2` |
|
| 637. |
During the first `18 min` of a `60 min` trip, a car has an average speed of `11 ms^-1.` What should be the average speed for remaining `42 min` so that car is having an average speed of `21 ms^-1` for the entire trip?A. `25.3 ms^-1`B. `29.2 ms^-1`C. `31 ms^-1`D. `35.6 ms^-1` |
|
Answer» Correct Answer - A `v_(av)=d/t=(d_1+d_2)/(t_1+t_2)` `21=((18)(11)+(42)(v))/60` `rArr v=25.3m//s` |
|
| 638. |
A particle is projected from the ground with an initial speed of v at an angle `theta` with horizontal. The average velocity of the particle between its point of projection and highest point of trajectroy is :A. `v/2sqrt(1+2cos^(2) theta)`B. `v/2 sqrt(1+cos^(2) theta)`C. `v/2 sqrt(1+3cos^(2)theta)`D. `v cos theta` |
|
Answer» Correct Answer - C |
|
| 639. |
A particle moves along the x-direction with constant acceleration. The displacement, measured from a convenient position, is 2 m at time `t= 0` and is zero when `t= 10 s.` If the velocity of the particle is momentary zero when `t = 6 s,` determine the acceleration a and the velocity v when `t= 10s.` |
|
Answer» Correct Answer - A::B `s=s_0+ut+1/2at^2` at `t=0, s=s_0=2m …(i)` at `t=10s,` `s=0=s_0+ut+1/2at^2` or `10u+50a=-2….(ii)` `v=u+at` `:. 0=u+at` `:. 0=u+6a (at t=6s)` `:. u+6a=0….(iii)` Solving Eqs. (ii) and (iii) we get, `u=-1.2 m//s` and `a=0.2 m//s^2` Now, `v=u+at` `:. v=(-12)+(0.2)(10)` `=0.8 m//s` |
|
| 640. |
At time `t= 0,` a particle is at `(2m, 4m).` It starts moving towards positive x-axis with constant acceleration `2 m//s^2` (initial velocity=0). After 2 s, an additional acceleration of `4 m//s^2` starts acting on the particle in negative y-direction also. Find after next 2 s. (a) velocity and (b) coordinates of particle. |
|
Answer» Correct Answer - A::B::D After 2s `v_1=u+a_1t_1` `=0+(2 hati)(2)=(4 hati) m//s` `r_1=1_i+1/2a_1t_1^2` `=(2 hati+4 hatj)+1/2(2hati)(2)^2` `=(6 hati+4 hatj)m` After next 2s (a) `v_2=v_1+a_2t_2` `=(4 hati)+(2hati-4hatj)(2)` `=(8hati-8 hatj)m//s` (b)`r_2=r_1+v_1t_2+1/2a_2t_2^2` `=(6 hati+4 hatj)+(4 hati)(2)+1/2(2 hati-4 hatj)(2)^2` `=(18 hati-4 hatj)m ` `:. ` Co-ordinates are, `x=18 m` and `y=-4 m` |
|
| 641. |
A man swimming in a river from a point. A on one bank has to reach a point C on other bank, which is at a distance `l` from the point B, directly opposite to A on other bank. River width is `d` and the current velocity is `u_(0)`. Find the minimum speed of swimmer relative to still water with which he should swim. |
|
Answer» Correct Answer - `[(u_(0)d)/(sqrt(l^(2) + d^(2))]` |
|
| 642. |
Read each statement below carefully and state with reasons, with it is true or false : (a) The magnitude of vector is always a scalar. ltBrgt (b) Each component of a vector is alwarys a scalar. (c ) The total path length is always equal to the magnitude of the displacement vector of a particle. (d) The average speed of a particle (defined as total path length diveked by the time taken to cover the path ) is eigther greater or equal to the magnitude of average velocity of the particle over the same interval of time. ltbRgt (e ) three vectors not lying in a planc ecan never add up to give a null vector. |
|
Answer» True , because magnitude is a pure nuber. (b) False , each component of a vector is also a vector. (c ) True only if the particle moves along a straight line in the sam direction, otherwise false. (d) True , because the total path length is either greater than or equal to the magnitude of the magnitude of the displacement vector. (e) True , as they can not be represented by the three sides of a triangle taken in the same order. |
|
| 643. |
A water sprinkler is positioned at O on horizontal ground. It issues water drops in every possible direction with fixed speed u. This way the sprinkler is able to completely wet a circular area of the ground (see fig). A horizontal wind starts blowing at a speed of `(u)/(2sqrt(2))`.Mark the area on theground that the sprinkler will now be able to wet. |
|
Answer» Correct Answer - A circle of same size shifted from the original circle by `DeltaX = (u^(2))/(2g)` in the direction of wind. |
|
| 644. |
A stone is projected from point `P` on the inclined plane with velocity `v_(0) = 10 m//sec` directed perpendicular to the plane. The time taken (in second) by the stone to strike the horizontal ground `S` is (Given `PO = l = 10`meter)(Take `g = 10m//s^(2)`) |
|
Answer» Correct Answer - 2 |
|
| 645. |
A stone is thrown from the top of a tower of height `h=10m` with speed `v-10m//s`. The distance of the landing point on the ground from the foot of the tower is `Rle2sqrt(3)k` in meter. Calculate `k`. Take `g=10m//s^(2)` |
|
Answer» Correct Answer - 5 |
|
| 646. |
Taxis leave station `X` for station `Y` every `10 min`. Simultaneously, a taxi also leaves station `Y` for station`X` every `10 min`. The taxis move at the same constant speed and go from `X` and `Y` or vice-versa in `2 h`, How many taxis coming from the other side will meet each taxi enroute from `Y` and `X`?.A. (a) ` 11`B. (b) `12`C. (c ) ` 23`D. (d) `24` |
|
Answer» Correct Answer - C No. of taxies leaving a station (A) in (2) hours ` = 2 xx 60 /10 = 12 `. Therefore a taxi standing at a location (B), will encounier ` 12 1 taxies in `2` hours. If taxi leaves station (B) and moves towards (A) will be meeting (11) more taxies enroute and it will be meeting the twelfth taxi at the location of (a) which is just going to start. Therefore this taxi is not counited in the total no. of taxies enroute . Thus the total no. of taxies encountered enroute ,brgt `12 + 11 23`. |
|
| 647. |
Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s)A. (a) ` vec v _(av)= 1/2 [ vec v (t_1) + vec v (t_2)]`B. (b) ` vec v = ( vec r(t_2)- vecr (t_1))/(t_2-t_1)`C. (c ) ` vec r = 1/2 ( vec v(t_2) - vec v (t_1)) (t_2-t_1)`D. (d) ` a_(av) = (vec v(t_2)- vec v(t_1))/(t_2-t_1)` |
|
Answer» Correct Answer - A::C For the motion of a particle if acceleration is not uniform then the relations ` vec v_(av) = 1/2 [ vec v(t-1) + vec v(t_2)] `si incorrect. ` vec v_(av) = (vec r(t_2) - vec r9t-1))/( (t_2-t_1)) ` is correct`. ` vec r = 1/2 [ vec v9t_2) - vec v( t-1) ] (t-2 -t_1) ` is incorrect, and ` vec a(av) = (vec v9t_1) - vec v(t_1))/(t_1 -t_1) ` is correct. |
|
| 648. |
Two shells are fired from a canon with seped (u) each, at angles if ` alpha` and ` beta` respectively with the horizontal . The time intrval between the shots is (t). They collide in mid air after time (T) from the first shot . Which of the following consitions must be satisfied ?A. ` alpha gt beta`B. ` T cos alpha = (T- t) cos beta`C. ` T- t) cos alpha = T cos beta`D. ` usin alpha T - 1/2 g T^2 = u sin bwta (T- t) - 1/2 g (T- t)^2` |
|
Answer» Correct Answer - A::B::D Since the first shot must have travelled for more time than the second shot before collinding in mid air. It will be so if the angle of projection of first short ` alpha` is greater than the angle of projection of second shot ` beta i.e. alpha gt beta `. At the colliding position of two shots , the borizontal distance (x) covereed by each shot is same so ` x= u cos alpha T = u cos beta (T - t)` or ` T cos alpha = (T-t) cos beta` At the colliding position of two shots, the vertical distance (y) covered by first shot is ` y= u sin alpha t - 1/2 g T^2` The verrtical distance `y` covered by secnd shot is ` y = u sin beta (T- t) - 1/2 g (T -t)^2` :. `u sin alpha T - 1/2 g T^2 = u sin beta (T-t) - 1/2 g (T-t)^2` . |
|
| 649. |
Explain the term Velocity-time graph. |
|
Answer» It is a graph of time versus velocity. Its slope t any point gives the acceleration at the corresponding instant. Distance covered in time t equals area under the velocity –time graph bounded by the time-axis. (i) For uniform motion, the velocity-time graph is a straight line parallel to the time-axis. (ii) For uniform acceleration, the velocity-time graph is a straight line inclined to the time-axis. (iii) For variable acceleration, the velocity time graph is a curve. |
|
| 650. |
A stone is thrown by a student from the bottom of a hill with a velocity `30 ms^(-1)` making an angle of `60^(@)` with the horizontal. If the slpe of the hill is `30 ^(@)` with the horizontal. Find the distance from the student to a point at which the stone fals on hill, use `g= 10 ms^(-2)`. |
|
Answer» Refr to Art. 2 (d) .9, the distance coverd on slope is `R= (2 u^(2) cos theta sin (theta-theta_(0)))/(g cos^(2) theta_(0))` Here, `u =30 ms^(-1)` `theta =60^(@), theta =30^(@), g=10 ms^(-2)` :. `R =(2 xx 30^(2) cos 60^(@) sin 9 60^(@) -30^(@)))/(10 xx cos ^(2) 30(#))` `=(2 xx 900 xx (1//2) xx (1//2))/(10 xx (sqrt3//2)^(2)` `=69 m`. |
|