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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.
| 851. |
The acceleration versus time graph of a particle moving in a straight line is show in. The velocity-time graph of the particle would be .A. `A straight lineB. A parabolaC. A circleD. An ellipse |
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Answer» Correct Answer - B `a=-(4)/(2) t+4 rArr a=-2t+4` `v=int a dt+C =int (-2t +4) dt + C=- t^(2) +4t+C` Hence, graph will be parablic. |
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| 852. |
Ship `A` is travelling with a velocity of `5 km h^-1` due east. A second ship is heading `30^@` east of north. What should be the speed of second ship if it is to remain always due north with respect to the first ship ?A. `10 km h^-1`B. `9 km h^-1`C. `8 km h^-1`D. `7 km h^-1` |
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Answer» Correct Answer - A (a) For B always to be north of `A`, the velocity components of both along east should be same. `v_2 cos 60^@ = v_1` `rArr v_2 = 10 km h^-1`. |
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| 853. |
A particle is projected up from a point at an angle of `30^(@)` with the horizontal. At any time `t`if `p=` linear momentum `y=` vertical displacement `x=` horizontal displacement then the kinetic energy `(K)` of the particle plotted against these parametes can beA. B. C. D. |
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Answer» Correct Answer - B::C::D |
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| 854. |
In a projectile motion let `v_(x)` and `v_(y)` are the horizontal and vertical components of velocity at any time `t` and `x` and `y` are displacements along horizontal and vertical from the point of projection at any time `t`.ThenA. `v_(y)-` graph is a straight line with negative slope and positive intercept.B. `x-t` graph is a straight line passing through originC. `y-t` graph is a straight line passing through originD. `v_(x)-t` graph is a straight line |
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Answer» Correct Answer - A::B::D |
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| 855. |
A projectile has the same range `R` for two angles of projections but same speed. If `T_(1)` and `T_(2)` be the times of flight in the two cases, then Here `theta` is the angle of projection corresponding to `T_(1)`.A. `T_(1)T_(2) prop R`B. `T_(1)T_(2)propR^(2)`C. `(T_(1))/(T_(2))=tantheta`D. `(T_(1))/(T_(2))=cot theta` |
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Answer» Correct Answer - A::C |
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| 856. |
Choose the correct alternative `(s)`A. If the greatest height to which a man can throw a stone is `h` then the greatest horizontal distance upto which he can throw the stone is `2h`B. The angle of projection for a projectile motioin whose range `R` is `n` times the maximum height `H` is `tan^(-1)(4//n)`C. The time of flight `T` and the horizontal range `R` of a projectile are connected by the equation `gT^(2)=2R tan theta` where `theta` is the angle of projection.D. A ball is thrown vertically up. Another ball is thrown at an angle `theta` with he vertical. Both of them remain in air for the same period of time. Then the ratio of heights attained by the two balls is `1:1` |
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Answer» Correct Answer - A::B::C::D |
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| 857. |
A particle with positive acceleration is slwing down if its ……………………………………is negative. |
| Answer» Correct Answer - initial velocity | |
| 858. |
Those vectors which have a starting point or a point of application are called …………………….. . |
| Answer» Correct Answer - polar vectors | |
| 859. |
A body is moving in a circular path with a constant speed. It has .A. A constant velocityB. A constant accelerationC. An acceleration of constant magnitudeD. An acceleration which varies with time in magnitude |
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Answer» Correct Answer - C ( c) `a_c = (v^2)/(r ) rarr` Constant in magnitude if `v` is constant. |
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| 860. |
The velocities of A and B shown in fig. Find the speed (in `ms^(-1)`) of block C. (Assume that the pulleys and string are ideal). |
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Answer» Correct Answer - `5m//s` By constrained equaiton for length `l_(1) +l_(2) +l_(3) +l_(4) =` constant `(dl_(1))/(dt) +(dl_(2))/(dt) +(dl_(3))/(dt) +(dl_(4))/(dt) = 0 …(1)` since `(dl_(1))/(dt) = (dl_(2))/(dt) = v_(A) = v_(B) = 1-3 =- 2m//s & (dl_(3))/(dt) =0` putting the value in eq.(1) `-2 +(-2) +0+(dl_(4))/(dt) = 0` `{:((dl_(4))/(dT)=4m//s,,rArr,v_(CB)=4m//s,),(vecv_(CB)=vecv_(C)-vecv_(B),,rArr,vecv_(C)=vecv_(CB)+vecv_(B),):}` `v_(C) = sqrt(v_(CB)^(2)+v_(B)^(2)) rArr v_(C) = sqrt(3^(2) +4^(2))` `rArr v_(C) = 5m//s` |
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| 861. |
A body is moving in a circle with a speed of `1 ms^-1`. This speed increases at a constant rate of `2 ms^-1` every second. Assume that the radius of the circle described is `25 m`. The total accleration of the body after `2 s` is.A. `2 m s^-2`B. `25 ms^-2`C. `sqrt(5) ms^-2`D. `sqrt(7) ms^-2` |
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Answer» Correct Answer - C ( c) `a_t = 2 ms^-2, v = u + a_t t = 1 + 2 xx 2 = 5 ms^-1` `a_c = (v^2)/( r) = (5^2)/(25) = 1 ms^-2` Net acceleration =`(sqrt(a_c^2 + a_t^2)) = (sqrt(1^2 + 2^2)) = (sqrt(5)) ms^-2` |
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| 862. |
A projectile can have the same range` R` for two angle of projection . If ` t-1` and ` t_2` be the terms of flight in the two cased then the initial velocity of projectile is?A. `1/4 gt_1t_2`B. `1/2 gt_1t_2`C. ` 1/2 g (t_1 +t_2)^2`D. `1/2 g (t_`1^2 +t_2^2)^(1/2)` |
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Answer» Correct Answer - D Horizontal range is same whe angle fo projection is ` alpha ` or ` (90^2- alpha)` . ltBrgt :. R= (u^2 sin 2 alpha)/g , ` t_1 = ( 2 u sin alpha)/g …(i) and ` t_2 = (2 u sin ( 90^@ -alpha))/g = ( 2 u cos alpha)/g ` …(ii) ` From (i), ` isn alpha = gt_1//2 u,` From (ii) , cos alpha = gt _2//2 u` :. sin ^2 alpha + cos^2 alpha= ( g/(2u)) ^2 (t_1^1 _t_1^2)` or ` u = g/2 (t_1^2 +t_1^2) ^(1//2)`. |
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| 863. |
What are the Relative velocity ? |
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Answer» The relative velocity of an object B with respect to object A when both are in motion is the rate of change of position of object B with respect to object A. Relative velocity of object B w.r.t. object A, vBA = vB - vA Relative velocity of object A w.r.t. object B, vAB = vA - vB |
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| 864. |
Define the term Motion under gravity. |
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Answer» When a body falls freely under the action of gravity, its velocity increases and the value of g is taken positive. When a body thrown vertically upward, its velocity decreases and the value of g is taken negative. |
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| 865. |
What is means by Free fall ? |
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Answer» In the absence of air resistance, all bodies fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small height is called free fall. The acceleration with which a body falls is called acceleration due to gravity and is denoted by g. Near the surface of the earth, g = 9.8 ms-2. |
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| 866. |
Define Positive acceleration and Negative acceleration. |
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Answer» Positive acceleration: If the velocity of an object increases with time, its acceleration is positive. Negative acceleration: If the velocity of an object decreases with time, its acceleration is negative. Negative acceleration is called retardation deceleration. |
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| 867. |
Sir i want to know that we have been taught about bodies going up in the air with some aceleration. so while solving questions why we dont consider acc. due to gravity...like if a body is going up with acceleration 'a' then shouldn't it's net acceleration be a-g? |
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Answer» The acceleration given in the question is actually the net acceleration (i.e considering g also) so we don't have to consider it separately. I hope this clears your query! |
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| 868. |
A car trvelling at `108 km h^(-1)` has its speed reduced to `36 km h^(-1)` after traelling a distance of `2000 m` Find the retardation (assumed uniform) and time taken for this process. |
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Answer» `u=108 km h^(-1) =30 ms^(-1), v=36 km h^(-1) =10 m s^(-1)` `v^(2)=u^(2)+2as` `rArr 10^(2) =30^(2)+ 2axx200 rArr a=-2 m s^(-2)` `v=u+at rArr 10=30-2t rArr t=10 s`. |
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| 869. |
A body moves at a speed of `100 ms^(-1)` for `10 s` and then moves at a speed of `200 m s^(-1)` for `20 s` along the same direction. The average speed is ……………………………., |
| Answer» `v_(av)=(s)/(t) =(100xx10+200xx20)/(10+20) =(500)/(3) m s^(-1)`. | |
| 870. |
In the arrangement shown in the figure, the block C begins to move down at a constant speed of `7.5 cm//s` at time `t = 0`. At the same instant block A is made to start moving down at constant acceleration. It starts at M and its speed is `30 cm//s` when it reaches `N (MN = 20 cm)`. Assuming that B started from rest, find its position, velocity and acceleration when block A reaches N. |
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Answer» Correct Answer - Position; 40 cm up from starting position `V_(B) = 45 cm//s (uarr)` `a_(B) = 22/5 cm//s^(2) (uarr)` |
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| 871. |
An electric fan has blades of length `30 cm` as measured from the axis of rotation. If the fan is rotating at `1200 rpm`, find the acceleration of a point on the tip of a blade. |
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Answer» Correct Answer - `480 pi^2 m s^-2` `omega = 1200 rpm = 1200 xx (2 pi)/(60) = 40 pi rad s^-1` `a_c = omega^2 r = (40 pi)^2 xx (30)/(100) = 480 pi^2 ms^-2`. |
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| 872. |
A point traverses half a circle of radius R during the interval `tau`.Calculate the following quantities averaged over that time (a)the mean velocity that is speed ltvgt , (b)the modulus of the mean vector lt v gt , ( c)the modulus of the average acceleration a if its tangential acceleration is constant. |
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Answer» Correct Answer - < v > =`(piR)/tau`,`l lt vevugtl=(2R)/tau , (2piR)/(tau^2)` |
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| 873. |
Two particles (A) and (B) are connected by a rigid rod ` AB`. The rod slides along perpendicular rails as shown here. The velocity of (A) to the left is ` 10 m//s`. What is the velocity of (B) ange ` alpha = 30^@`. .A. (a) ` 9.8 m//s`B. (b) ` 10 m//s`C. ( 5.8 m//s`D. ( `17 .3 m//s` |
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Answer» Correct Answer - D Let particle (B) move upwards with velocity (v). At an instant. Let ` OA = x ` and ` OB = y, Ab =l `. Then ltbRgt ` X^2 + y^2 = l^@` Differentating it with time (t) we have ` 2 x (dx0/(dt) + 2 y (dy)/(dt) =0` `(dy)/(dt)=- x/y (dx)/(dt) =- (dx //dt)/(tan 30^2) =- 10 xx sqrt 3` ` =- 17. 32 m//s`. |
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| 874. |
The coordinates of a moving particle at any time `t` are given by `x = alpha t^(3)` and `y = beta t^(3)`. The speed of the particle at time `t` is given byA. `3tsqrt(alpha^(2) +beta^(2))`B. `3t^(2) sqrt(alpha^(2) +beta^(2))`C. `t^(2) sqrt(alpha^(2) +beta^(2))`D. `sqrt(alpha^(2) +beta^(2))` |
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Answer» Correct Answer - B `x = alphat^(3), y = betat^(3)` `V_(x) = (dx)/(dt) = 3 alphat^(2)` `V_(y) = (dy)/(dt) = 3betat^(2)` Resultant velocity `V = sqrt(v_(x)^(2) +v_(y)^(2))` `= sqrt(9alpha^(2)t^(4) + 9 beta^(2)t^(4))` `= 3t^(2) sqrt(alpha^(2) +beta^(2))` |
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| 875. |
A body is moved along a straight line by a machine delivering constant power . The distance moved by the body is time `t` is proptional toA. `t^(3//4)`B. `t^(3//2)`C. `t^(1//4)`D. `t^(1//2)` |
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Answer» Correct Answer - B `P = Fv` `P = mav` `P = m(dv)/(dt).v` `int vdv = int(P)/(m)dt` `v^(2) prop t` `v prop t^(1//2)` `(dx)/(dt) prop t^(1//3)` `dx prop t^(1//3) dt` `x prop t^(2//3)` |
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| 876. |
A graph of x versus t is shown in figure. Choose correct alternative from below.A. (a) The particle was released from rest at ` t= 0`B. (b) At (B) the acceleration ` a gt 0`C. (c ) At (C ), the velocity and the acceleration vanish .D. (d) ` Average velocity for the motin between (A) and (D) is positive |
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Answer» Correct Answer - A::C From graph , when ` t-0`, the particle is released from rest at (A), hence ` v=0` As, (B) the graph is parallel to time axis hence velocity is constant there. Thus accleration (A) is zero. AT (C ), the graph changes slope, where velocity and accelerationn vanish. ltbRgt Average velocity for motion between (A) and (D) is negatve because the value of (x) is dectrasing with tiem (t0. The slope of graph (which represents speed) is more at (D) than at (E). Therefore, speed at (D) is more than that at (E)`. |
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| 877. |
A particle of mass m moves on positive x-axis under the influence of force acting towards the origin given by `-kx^2 hat i.` If the particle starts from rest at `x=a,` the speed it will attain when it crosses the origin isA. `(sqrtk/(ma))`B. `(sqrt2k/(ma))`C. `(sqrt(ma)/2k)`D. None of these |
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Answer» Correct Answer - D `a=F/m=(-kx^2)/m` `:. v.(dv)/(dx)=-(kx^2)/m` or `int_0^v vdv=int_a^0-(kx^2)/m dx` `v^2/2=(ka^3)/(3m)` `:. v=(2ka^3)/(3m)` |
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| 878. |
What is meant by Horizontal Projectile ? |
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Answer» A body be projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant. |
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| 879. |
What is Circular motion and its types ? |
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Answer» Circular motion is another example of motion in two dimensions. To create circular motion in a body it must be given some initial velocity and a force must then act on the body which is always directed at right angles to instantaneous velocity. Circular motion can be classified into two types : Uniform circular motion and non-uniform circular motion. |
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| 880. |
How can Work done by Centripetal Force? Give an examples. |
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Answer» The work done by centripetal force is always zero as it is perpendicular to velocity and hence instantaneous displacement. Example : (i) When an electron revolve around the nucleus in hydrogen atom in a particular orbit, it neither absorb nor emit any energy means its energy remains constant. (ii) When a satellite established once in a orbit around the earth and it starts revolving with particular speed, then no fuel is required for its circular motion |
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| 881. |
Define Centrifugal Force. |
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Answer» It is an imaginary force due to incorporated effects of inertia. Centrifugal force is a fictitious force which has significance only in a rotating frame of reference. |
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| 882. |
Give the different situation of Centripetal force. |
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| 883. |
Define Time period (T) of Circular Motion. |
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Answer» Time period (T) : In circular motion, the time period is defined as the time taken by the object to complete one revolution on its circular path. |
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| 884. |
Two vectors ` vec A and vec B` are added. Prove that the magitude of the resultant v ector can bot be greater than ( A +B) and smaller than (A - B) or (B -A). |
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Answer» The magnitude of resultant vector ` vec R` of two vectors ` vec Aand vec B` is given by ` R =sqrt ( A^2 + B^2 + 2 AB cos theta ) ` Case 9i). `R` will be maximum if ` cos theta =1 or theta =180^@` then ` R =sqrt( A^2 + B^2 + 2 AB (1))` = `(A = B) or (B -A)`. |
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| 885. |
Are the commutative law and associtive law applicable to vectors subtraction. |
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Answer» We cannot apply the commutative law to the vectors subtraction because ` vec A -vec B != vec B - vec A`. As ` (vec A -vec B) and (vec B - vec A) are acting in opposite directions. The associative laws to vectors subtraction cannot be applied because ` ( vec A- vec B) - vec C != vec A - (vec B- vec C)`. |
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| 886. |
A golfer standing on level ground hits a ball with a velocity of `52 m s^-1` at an angle `theta` above the horizontal. If `tan theta = 5//12`, then find the time for which then ball is atleast `15 m` above the ground `(take g = 10 m s^-2)`.A. 1 sB. 2 sC. 3 sD. 4 s |
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Answer» Correct Answer - B (b) Let at any time `t`, the ball be at height of `15 m`. `S_y = u_y t + (1)/(2) a_y t^2` `rArr 15 = u sin prop t - (1)/(2) "gt"^2` `rArr 15 = 52 xx (5)/(13) t - (1)/(2) xx 10 t^2` `rArr t^2 - 4t + 3 = 0 rArr (t -1)(t - 3) = 0` `rArr t = 1 s, t = 3 s`. Required time is `3 - 1 = 2 s`. |
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| 887. |
A ball is projected from the floor of a long hall having a roof height of `H = 10 m`. The ball is projected with a velocity of `u = 25 ms^(–1)` making an angle of `theta = 37^(@)` to the horizontal. On hitting the roof the ball loses its entire vertical component of velocity but there is no change in the horizontal component of its velocity. The ball was projected from point A and its hits the floor at B. Find distance AB. |
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Answer» Correct Answer - `20 (1+ sqrt(2)) m` |
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| 888. |
A particleAmoves with velocity(2ˆi−3ˆj)m/sfrom a point(4,5m)m. At the same instant a particleB, moving in the same plane with velocity(4ˆi+ˆj)m/spasses through a pointC(0,−3)m. Find thex-coordinate (inm) of the point where the particles collide. |
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Answer» Correct Answer - 8 If particle collide sfter time t at point P, then position of both after time t should be same both have uniform motion. by `S=S_(0)+vxxt` for `I^(st)` particle `S_(I)=4hat(i)+5hat(j)+(2hat(i)-3hat(j))xxt` for `II^(nd)` particle `S_(II)=-3hat(j)+(4hat(i)+hat(j))xxt` but `S_(I)=S_(II)` `4hat(i)+5hat(j)+(2hat(i)-3hat(j))t=-3+(4hat(i)+hat(j))+(4hat(i)+hat(j))xxt` `4hat(i)+8hat(j)=2hat(i)t+4hat(i)t` by comrision along `xxDeltay` `t=2` sec `S_(I)=4hat(i)+5hat(j)+(2hat(i)-3hat(j))2` `=8hat(i)-hat(j)` `x=8 m` |
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| 889. |
A particle is projected with velocity `30^(@)` above on an inclined plane, inclination of which from horizontal is `37^(@)`. Choose the appropriate path (air resistance is negligible)A. B. C. D. |
| Answer» Correct Answer - D | |
| 890. |
Two cars are moving in the same direction with the same speed `30 km//hr`. They are separated by a distance of `5 km`, the speed of a car moving in the opposite direction of it meets these two cars at an interval of `4` minutes, will be.A. 30 kphB. 35 kphC. 40 kphD. 45 kph |
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Answer» Correct Answer - D |
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| 891. |
Give examles of variable velocity, when (a) the magnitude is constant and the direction is changingn (b) the manitude is cahnging and the driection remains the same (c) both the magnitude and direction are changing |
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Answer» (i) A particle executing circualr motion with uniform speed (ii) A freely falling body or a body projected up. (iii) Conider a vehicle moving on a busy road. |
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| 892. |
What is circular motion? Give example |
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Answer» Circular motion is defined as a motion described by an object traversing a circular path. Example – The whirling motion of a stone attached to a string |
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| 893. |
What is rotational motion? Give example. |
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Answer» During a motion every point in the object traverses a circular path about an axis except the points located on the axis, is called as rotational motion. Example – Spinning of the earth about its own axis. |
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| 894. |
What is vibratory motion? Give example. |
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Answer» If an object or particle executes a to and fro motion about a fixed point, it is said to be in vibratory motion. Example – Vibration of a string on a guitar. |
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| 895. |
The dimension of point mass is – (a) 0 (b) 1 (c) 2 (d) kg |
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Answer» Correct answer is (a) 0 |
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| 896. |
What is one dimensional motion? Give example. |
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Answer» One dimensional motion is the motion of a particle moving along a straight line. Example – Motion of a train along a straight railway track. |
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| 897. |
What is three dimensional motion? Give example. |
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Answer» If a particle moving in used three dimensional space, then the particle is said to be in three dimensional motion. E.g. A bird flying in the sky. |
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| 898. |
What is two dimensional motion? Give example. |
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Answer» If a particle moving along a curved path in a plane, then it is said to be in two dimensional motion. Example – Motion of a coin on a carrom board. |
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| 899. |
Will two dimensional motion with an acceleration only in one dimension? |
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Answer» Yes. In oblique projection, the acceleration is acting vertically downward but the object follows a parabolic path. |
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| 900. |
Explain with examples of motion in one, two and three dimensions. |
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Answer» Three types of dimensions motion : One dimensional :
e.g., Motion of car on a straight road. Motion of freely falling body. Two dimensional :
e.g., Motion of car on a circular turn. Motion of billiards ball. Three dimensional :
e.g., Motion of flying kite. Motion of flying insect. |
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