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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.
| 901. |
What are the types of motion? |
Answer»
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| 902. |
What is frame of reference? |
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Answer» In a coordinate system, the position of an object is described relative to it, then such a coordinate system is called as frame of reference. |
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| 903. |
Define Frame of Reference. |
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Answer» It is a system to which a set of coordinates are attached and with reference to which observer describes any event. Rest and motion are relative terms. It depends upon the frame of references. |
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| 904. |
What are the Rest and Motion ? |
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Answer» If a body does not change its position as time passes with respect to frame of reference, it is said to be at rest. And if a body changes its position as time passes with respect to frame of reference, it is said to be in motion. |
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| 905. |
The speed of a projectile (u) rekuces by ` 50 %` on reachig maximum hight. What is the range on the horizontal plane ? |
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Answer» If ` theta` is the angle of projection, then velocity of projectile at height point `= u cos theta` As per question, ` u cos theta = ( 50)/( 100) u = 1/ 2 u ` or ` cos theta = 1/2 = cos 60^@` or ` theta =60^@` Horizontal range, `R= ( u^2 sin 2 theta)/g = u^2 /g sin 2 xx 60^@ = u^2 /g sin 120^@ = u^2/g xx sin ( 180^@- 60^@)` ` u^2/g sin 60^@ = u^@/g 60^2 = u^2/g xx (sqrt 3)/2`. |
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| 906. |
How much hight above the ground a mancan throw a ball if he is able to throw the same ball upto maximum distance of `60 m? |
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Answer» Her, `R _(max)=60 m`. As, `R_(max) =u^(2) /g , so 60 =u^(2)/g` …(i) Let (h) be the maximum height attained by the ball. Taking vertical upward motion of ball upto maximum height (h), we have `u&(y) =u, a)(y) =- g, u_(y) =0, y=0` As v_(y)^(2) =u_(y)^(2) +2 a)(y) y` :. 0=u^(2) +2 (-g) h` or `h=u^(2)/(2 g) =1/2 xx 60 =30 m`. |
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| 907. |
Statement-I : Graph (1) represent one dimensional motion of a particle. While graph (2) can not represent 1-D motion of the particle. (here x is position and t is time) Statement-II : Particle can have only one position at an instant.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 | |
| 908. |
Figure (a) and (b) shows a system of two masses A and B a motor M. find the relation in velocities of mass A and B, if the motor winding speed is v. |
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Answer» Correct Answer - [(a) `2v_(B) + v_(A) = v(v_(A) rarr; v_(A) rarr)` (b) `4v_(B) + v_(A) = v(v_(B) uarr; v_(A) uarr)`] |
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| 909. |
Figure shows a pulley over which is string passes and connected to two masses A and B. Pulley moves up with a velocity `V_(P)` and mass B is also going up at a velocity `V_(B)`. Find the velocity of mass A if (a) `V_(P)=5ms^(-1)` and `V_(B)=10ms^(-1)` (b) `V_(P)=5ms^(-1)` and `V_(B)=-20 ms_(-1)`. |
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Answer» Correct Answer - [(a) 0 m/s, (b) 30 m/s] |
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| 910. |
In the figure shown, acceleration of 1 is x (upwards). Acceleration of pulley `P_(3)`, w.r.t. pulley `P_(2)` is y (downwards) and acceleration of 4 w.r.t. to pulley `P_(3)` is z (upwards). Taking upward +ve and downward -ve then `{:(,"Column I",,,"Column II"),((A),"Absolute acceleration of 2",,(p),(y-x) "downwards"),((B),"Absolute acceleration of 3",,(q),(z-x-y) "upwards"),((C),"Absolute acceleration of 4",,(r),(x+y+z) "downwards"),(,,,(s),"None"):}` |
| Answer» Correct Answer - (A)-s, (B)-r, (C)-q | |
| 911. |
Find the relation in the accelerations of the three masses shown in fig. |
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Answer» Correct Answer - [(a) `a_(A) + 2a_(B) + 2a_(C) = 0(a_(A) darr; a_(B) darr; a_(C) darr)`, (b) `2a_(A) + a_(B) + 2a_(C) = 0(a_(A) darr; a_(B) darr; a_(C) darr)`] |
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| 912. |
The `v-s` and `v^(2)-s` graph are given for two particles. Find the accelerations of the particles at `s=0`. . |
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Answer» `a_(1)=(vdv)/(ds)`, where `v=20 m s^(-1)` (at s=0) and `(dv)/(ds)=-(2)/(5)=(a_(1))/(v) =a_(1)/(20)` Then, `a_(1) =-8 m s^(-2)` `a_(2)=(1)/(2)(d(v^(2)))/(ds)`, where `(d(v^(2)))/(ds) =-(2)/(5)` Then, `a_(2)=-0.2 m s^(-2)`. |
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| 913. |
The acceleration versus time graph of a particle is shown in. The respective `v-t` graph of the particle is . .A. .B. .C. .D. . |
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Answer» Correct Answer - A From `0` to `t_(1)`. Acceleration is increasing linearly with time, hence, `v-t` graph should be parabolec upwards. From `t_(1`) to `t_(2)` acceleration is dereasing linearly with time, hence, the `v-t` graph should be parabolic downwards. |
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| 914. |
Two particles `A` and `B` are located in `x-y` plane at points `(0,0)` and `(0,4m)`. They simultaneoulsy start moving with velocities `v_(A)=2hatjm//s` and `v_(B)=2hatim//s`. Select the correct alternative(s)A. the distance between them is constantB. The distance between them first decreases and then increasesC. the shortest distance between them is `2sqrt(2)m`D. Time after which they are at minimum distances is `1s` |
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Answer» Correct Answer - B::C::D |
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| 915. |
Let `r` be the radius vector of a particle in motion about some reference point and `r` its modulus. Similarly, `v` be the velocity vector and `v` its modulus. ThenA. `v!=(dr)/(dt)`B. `v=(dr)/(dt)`C. `v=|(dr)/(dt)|`D. `|dr|!=dr` |
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Answer» Correct Answer - A::C::D |
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| 916. |
a particle moving along a straight line with uniform acceleration has velocities `7 m//s` at A and `17 m//s` at C. B is the mid point of AC. Then :-A. the average velocity betwene `R` and `O` is `15m//s`B. the ratio of time to go from `P` to `R` and that from `R` to `Q` is `3:2`C. the velocity at `R` is `10m//s`D. the average velocity between `P` and `R` is `10m//s` |
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Answer» Correct Answer - A::B::D |
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| 917. |
The coordinate of a particle moving in a plane are given by ` x(t) = a cos (pt) and y(t) = b sin (pt)` where `a,b (lt a)` and `P` are positive constants of appropriate dimensions . ThenA. the path of the particle is an ellipseB. The velocity and acceleration of the particle are normal to each other at `t=pi//2p`C. the acceleration of the particle is always directed towards a fixed pointD. the distance travelled by the particle in time interval `t=0` to `t=pi//2p` is a |
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Answer» Correct Answer - A::B::C |
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| 918. |
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 isA. `pi((u^2)/(g))^2`B. `pi((u^2)/(2 g))^2`C. `pi((u)/(g))^2`D. `pi((u)/(2 g))^2` |
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Answer» Correct Answer - A (a) A bullet fired at angle `45^@` will fall maximum away, and all other bullets will fall with this bullet fired at `45^@`. `R_(max) = (u^2)/(g)` Maximum area covered `= pi R_(max)^2 = pi ((u^2)/(g))^2`. |
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| 919. |
Which of the following statements is FALSE for a paricle moving in a circle with a constant angular sppeed?A. The velcoity vector is tangent to the circleB. The acceleration vector is tangent to the circleC. The acceleration vector point to the cnetre of the circleD. The velocity and acceleration vectors are perpendicular to each other |
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Answer» Correct Answer - B For a particle moving in a circle with constant angular speed, velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of circle or is always point towards the centre of circle or is always along radius of he circle. since, tangential vector is prependicular to raidal vector, therefore, velocity vector will be perpendicular to the acceleration vector. But is no case acceleration vector is tangent to the circle |
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| 920. |
If ` vec A + vec B = vec A- vec B`, then vec B` is a…………………………….. . |
| Answer» Correct Answer - null vector | |
| 921. |
A paricle is projected at an anged `theta` from the horizontal with kinetic energy (T). What is the kinetic energy of the particle at the highest point ? |
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Answer» At highest point, velocity of projectile K.E. of projectile at highest point `= 1/2 m (cos theta )^2 = (1/2 m u^2) cos^2 theta` ` = T cos^2 theta`. |
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| 922. |
(hat i+hat j+ hat k) makes an angle…………………………with each of X, Y and Z-axis. |
| Answer» Correct Answer - ` 54.74^@` | |
| 923. |
Assertion (A) : Quartz clocks are more accurate than pendulum clocks. Resson ( R) : Quartz clocks use periodic motion of simple pendulum for the measurement of time.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true, but R is not the correct explanation of A.C. A is true but R is falseD. Both A and R are false. |
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Answer» Correct Answer - C Quartz clocks contain an electric circuit and are more accurate than pendulum clocks. |
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| 924. |
The time taken by a simple pendulum to complete 20 oscillations is given in the table as shown below study the and write the following steps in sequential order to determine the frequency of its oscillations. (A). Find the reciprocal of average time period (T). (B). Find the ratio of time and number of oscillations. (C). Find the average of the values obtained in the previous step.A. BCAB. ABCC. ACBD. Both (a) and ( c) |
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Answer» Correct Answer - A Find the time period in each step. Time period = `("time taken for 20 oscillations")/("Number of oscillations")` Find the average value of the time period obtained in the percous step. Find the reciprocal of the average time period as frequency. |
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| 925. |
Assertion (A): When the length of a pendulum is 100 cm , the length used to suspend the bod is less than 100 cm. Reason ( R): The length of the pendulum is the distance between the point of suspension and the bottom most point.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true, but R is not the correct explanation of A.C. A is true but R is falseD. Both A and R are false. |
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Answer» Correct Answer - C For the length of the pendulum to be 100 cm, the length of the thread should be (100 -R) cm, where R iis the radius of the bob. |
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| 926. |
An object is moving along+ve x-axis with a uniform acceleration of 4 ms^(-2)`. At time ` t=0`. X= 4 m` and ` v=2 ms^(-1)``. (a) What will be the velocity and position of the object at time ` t=3 s? (b) What will be the position of the object when it has a velocity ` 8 ms^(-1)` ? |
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Answer» Here, ` x (0) = 4 m, v (0) = u = 2 ms^(-1)`, ` a = 4 ms^(-2)` (a) When ` t= 3 s,` ` v= u + at = 2 + 4 xx 3 14 am^(-1)` Ppsotion of object at time ` t=3 s`, ` x=x_0 + ut = 1/2 at^2 = 4 + 2 xx 3 + 1/2 xx 4 xx 3^2 = 28 m` (b) When ` u= 8 ms^(-1)`, posotion ` X (t) of the object is giben by ltbRgt ` `v^2 - u^2 2 a ( x- x_0)` or ` x=x_0 + ( v^2-u^2)/( 2a) = 4 + ( 8^2 - 2^2)/( 2 xx 4)` ` = 4 + 15 //2 = 11. 5 m`. |
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| 927. |
Assertion: A body having non zero acceleration can have a constant velocity. Reason: Acceleration is the rate of change of velocity.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true, but R is not the correct explanation of A.C. A is true but R is falseD. Both A and R are false. |
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Answer» Correct Answer - A Acceleration is the rate of change in velocity. Hence, it is zero for a body moving with constant velocity. |
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| 928. |
Assertion (A), If the speed of a car moving towards the east is ` 20 ms^(-1)` , its velcoity is `20 m s^(-1)` towards the east. Reason ( R): The velocity of a body is spend in a specified direaction.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true, but R is not the correct explanation of A.C. A is true but R is falseD. Both A and R are false. |
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Answer» Correct Answer - A Velocity has both magnitude and direaction its magnitude gives speed. Hene, both the statements are correct and R is the correct explanation of A. |
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| 929. |
A car starts from and attains a velocity of 54 km ` h^(-1)` if the acceleration of the car is from rest and attains a velcoity of 54 km `h^(-1)` . If the acceleration of the car is ` 3 m s^(-2)`, write the following steps in sequential order to find the time taken to accelerate the car. (A) Write the relation between u,v,a and t, where t is the time taken to accelerae the car. (B) Note the final velocity (v) and convert it into SI unit. (D) substitute the given values and obtain the vlaue of t using ` t= (v-u)/a`A. ABCDB. ABCDC. CADBD. CBAD |
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Answer» Correct Answer - C ( c) Note the intial velocity of the car u=0 and note down its acceleration. (B) Note th final velocity (v) and convert into SI unit. (A) Write the relation between u,v,a and t, where t is the time taken to accelerate the car . (D) Substitute the given values and obtain the valu of t , using `t = ( v-u)/a` |
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| 930. |
A motor boat covers the distance between the two spost on the tiver in `9 h and 13 h` down stream and upstream respectively. Find the time required by the boat to cover this distance in still water. |
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Answer» Let (u) be the velcity of boat in still water and (v) be the velcity of river. Let (x) be the distance between two given spets. As per question. For downstream, `u +v =x/9` For upstream, `u-v =x/(13)` On adding, `2u =x/9 +x/(13) =(22x)/(117) or u =(11x)/(117)` Time taken by boat to cover distance (x) in still water is `t=x/u = x /(11 x//117) =10.63 h`. |
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| 931. |
A jet plane is flying horizontally with a velocity `600 km h^(-1)`. The burnt gases are injecting from the rear of the get plane with the velocity `2000 km h^(-1)`. With respect to the jet plane. Find the velocity of burnt gases with respect to a person on the ground. |
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Answer» Let the motion along +ve x-direction be taken positive and motion alon-ve x-direction be negativ. Let the jet plane be moving along positive x-direction. The burnt gases will be moving along negatibe x-direction. Therefore, velocity of place, `v_(j) =600 km h^(-1)` velcoty of burnt gases with respect to jet plane `v_(gj) =v_(g) v_(j) =- 2000 km h^(-1)` or `v_(g) =- 2000 km h^(-1) +v_(j) =-2000 + 600 ` `=- 1400 km h^(-1)` The velcity of burnt gases w,r,t, person standing onthe ground `=(|v_(g) -v_(G) )= - 1400 -0 `= - 1400 km h^(-1)`. |
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| 932. |
Two balls of different masses (one lighter and other heaver) are thrown vertically upwards with the same speed. Which one will pass through the point of projection in the downward direction with greater speed? |
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Answer» Let (u) be the initial velocity of projection of body and (v) be the velocity of the same body while passing downwards through point of projection. The displacement of body ` S=0`. Using the relation ` v^(2) =u^(2) + 2 aS`, and `u=u, =? , =- g, S=0`, we have ` v^(2) =u^(2) + 2 (-g) xx 0 = u^(2) or `v=u` It means the final speed is independent of mass of the body. Hecnec, bothe the bodies will acquire the same speed while passing through point of projection. |
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| 933. |
A ball is thrown at different angles with the same speed `u` and from the same points and it has same range in both the cases. If `y_1 and y_2` be the heights attained in the two cases, then find the value of `y_1 + y_2`.A. `(u^2)/(g)`B. `(2 u^2)/(g)`C. `(u^2)/(2 g)`D. `(u^2)/(4 g)` |
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Answer» Correct Answer - C ( c) `y_1 = (u^2 sin^2 theta)/(2 g), y_2 =(u^2 sin^2(90^@ - theta))/(2 g) = (u^2 cos^2 theta)/(2 g)` `rArr y_1 + y_2 = (u^2)/(2 g)`. |
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| 934. |
A projectile can have same range `R` for two angles of projection. It `t_1 and t_2` are the times of flight in the two cases, then what is the product of two times of flight ?A. ` R`B. ` 1//R`C. ` R^2`D. ` 1//R^@` |
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Answer» Correct Answer - A Horizontal range is same when angle of projection is ` theta` or (90^2-theta)` ` R= ( u^2 sin 2 theta)/g , t_1 = ( 2 u sin theta )/g` and ` t_2 = (2 u sin (90^2 - theta)/g = ( 2 u cos theta)g` :.t_1 t_2 = ( 2 u sin theta)/g xx ( 2 u cos hteta)/g = ( 2 u^2 ( 2 sin theta cos theta)/g^2` `= ( 2 u^2 sin 2 theta )/g = 2/g ( (u ^2 sin 2 theta)/g)` or ` t_1 t-2 = 2/g R` :. t-1 t_2 prop R`. |
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| 935. |
The horizontal range of a projectile is `2 sqrt(3)` times its maximum height. Find the angle of projection. |
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Answer» If `u and prop` are the initial velocity of projection and angle of projection, respectively, then Maximum height attained `=(u^2 sin^2 prop)/(2 g)` Horizontal range `(2 u^2 sin prop cos prop)/(g)` According to the problem, `(2 u^2 sin prop cos prop)/(g) = 2 sqrt(3) ((u^2 sin^2 prop)/(2 g)) rArr tan prop = ((2)/(sqrt(3)))` `rArr prop = tan^-1 ((2)/(sqrt(3)))`. |
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| 936. |
A projectile can have same range `R` for two angles of projection. It `t_1 and t_2` are the times of flight in the two cases, then what is the product of two times of flight ?A. `t_1 t_2 prop R^2`B. `t_1 t_2 prop R`C. `t_1 t_2 prop (1)/( R)`D. `t_1 t_2 prop (1)/(R^2)` |
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Answer» Correct Answer - B (b) The horizontal range is the same for the angles of projection `theta and (90^@ - theta)`. `t_1 = (2 u sin theta)/(g), t_2 = (2 u sin (90^@ - theta))/(g) = (2 u cos theta)/(g)` `t_1 t_2 = (2 u sin theta)/(g) xx (2 u cos theta)/(g) = (2)/(g) [(u^2 sin 2 theta)/(g)] = (2)/(g) R` where `R = (u^2 sin 2 theta)/(g)` Hence, `t_1 t_2 prop R` (as R is constant). |
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| 937. |
A batsman hits a ball at an angle of `306^@` with an initial speed of `30 ms^-1`. Assuming that the ball travels in a verticle plane, calculate. (a) The time at which the ball reaches the highest point. (b) The maximum height reached. ( c) The horizontal range of the ball. (d) The time for which the ball is in the air. |
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Answer» Here `theta = 30^@ , u = 30 m s^-1` (a) The time taken by the ball to reach the highest point is half the total time of flight. As the time of ascending and decending is same for a projectile without air resistance, the time to reach the highest point `t_H = (T)/(2) = (u sin theta)/(g) = (30)/(10) xx sin 30^@ = 1.5 s` (b) The maximum height reached is `(u^2 sin^2 theta)/(2 g) = ((30)^2 xx (sin 30^@)^2)/(2 g) = (900)/(2 xx 10 xx 4) = 11.25 m` ( c) Horizontal range `= (u^2 sin 2 theta)/(g)` `=((30)^2 sin 2 (30^@))/(10)` `= (900sqrt(3))/(20)m = 45 sqrt(3) m` (d) The time for which the ball is in air is same as its time of flight, i.,e., `(2 u sin theta)/(g) = (2 xx 30 xx sin 30^@)/(10) = 3 s`. |
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| 938. |
An object thrown vertically. Thevelocity-time graph for the motion of the particle is .A. .B. .C. .D. . |
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Answer» Correct Answer - D At `t=0` velovity is positive and maximum. As the particle goes up velocity decreases and becomes zero at the highest point. When the particle starts coming down, velocity increases in the negative dierction. |
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| 939. |
The displacement-time graph of a body is shown in. . The velocity-time graph of the motion of the body will be .A. .B. .C. .D. . |
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Answer» Correct Answer - D From `0` to `t_(1)`, Verlocity id positive and constant as indicated by positive and constant slope. From `t_(1)` to `t_(3), slope is zero hence velocty is zero. From `t_(3)` to `t_(4)` velocity is negative and constant as indicated by negative and constant slope. Option `d` satisfies all these observations. |
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| 940. |
The slope of velocity-time graph gives – (a) velocity (b) acceleration (c) force (d) displacement |
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Answer» (b) acceleration |
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| 941. |
The slope of displacement-time graph gives – (a) velocity (b) acceleration (c) force (d) displacement |
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Answer» (a) velocity |
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| 942. |
The second derivative of position vector with respect to time is –(a) velocity (b) acceleration (c) force (d) displacement |
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Answer» (b) acceleration |
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| 943. |
The velocity-time graph of a body is given in. The maximum acceleration in `ms^(-1)` is . .A. `4`B. `3`C. `2`D. `1` |
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Answer» Correct Answer - A Maximum acceleration will be from `30` to `40 s`, because slope in this intervl is maximum. `a =(v_(2)-(v_(1))/(t_(2)-t_(1)) =(60-20)/(40-30) =4 m s^(-2)`. |
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| 944. |
The slope of the position – time graph will give – (a) displacement (b) velocity (c) acceleration (d) force |
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Answer» Correct answer is (d) force |
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| 945. |
What do you understand by positive an dbegative time. |
| Answer» The instant of time which is taken after the origin of time (i.e. zero) is called positive time and the instant of time which is taken before the origin of time is called begative time. | |
| 946. |
Statement I: The average velocity of the body may be equal to its instantaneous velocity. Statement II: For a given time interval of a given motion, average veocity is single valued while average speed can have many values.A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I.B. Statemnt I is true, Statement II is true, Statement II is true, Statement II is false.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
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Answer» Correct Answer - C The average velocity of the body may be equal tio tis instantaneous velocity, becoause instantaneus velocity can take any value. For a given time interval of a given motion, both average velocity and average speed can have only one value as displacement and distance will have single values. |
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| 947. |
Can position-time graph have negative slope ? |
| Answer» Yes, when the velocity of the obect is begative. | |
| 948. |
The distance (x) particle moving in one dismension, under the action of a constant force is related to time (t) by equation `tsqrt x +3` where (x) is in vettres and (t) in seconds. Find the displacement of the particle when its velcity is zero. |
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Answer» Given, `t= sqrt x+ 3 or sqrt x = t -3 ` Squaring both the sides, we get ` x=(t-3) ^(2) =t^(2) -6 t +9` Diferntiating it w.r.t. time (t) we get velocity, ` v=(dx)/(dt) =2 t-6` when `v=0` , then `2 t -6 =0 or t=3 second`. At ` t=3 `, distplacement is given by, ` x=t^(2) -6 t+ 9 = 3^(2) -6 xx 3+=0`. Hence, displacement of the particle is zero when its velocity is zero. |
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| 949. |
The velovity-time graph of a particle moving in a straitht line is shown in . The acceleration of the particle at `t=9 s` is. .A. ZeroB. `5 m s^(-2)`C. `-5 m s^(-2)`D. `-2 m s^(2)` |
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Answer» Correct Answer - C Acceleration between `8` and `10 s or at `t=9 s`: `a= (v_(2)-v_(1))/(t_(2) -t_(1)) =(5-15)/(10-8) =- ms^(-2)`. |
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| 950. |
The average velocity of a particles is equal to its displacement-time graph ? |
| Answer» When the average velocity of a particle is rqual to its instantaneous velcity, the particle is moving with a uniform velcity aling a straight line. The displacement-time graph of a unitorm motion is a straight line inclined to the time axis. | |