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51.

A fly wheel is rotating about its own axis at an angular velocity `11 rad s^(-1)`, its angular velocity in revolation per minute isA. `105`B. `210`C. `315`D. `420`

Answer» Correct Answer - A
`omega=2pin,n=omega/(2pi)(rps),n(rps)=60xxn("rpm")`
52.

A stationary wheel starts rotating about its own axis at uniform rate amgular acceleration `8rad//s^(2)`.The time taken by its to complete `77` rotation isA. `5.5 sec`B. `7 sec`C. `11 sec`D. `14 sec`

Answer» Correct Answer - C
`theta=2piN,t=sqrt((2theta)/(alpha))`
53.

Keeping the speed of projection constant, the angle of projection is increased from `0^(@)` to `90^(@)`. Then the horizontal range of the projectileA. goes on increasing up to `90^(@)`B. decreases up to `90^(@)`C. increases up to `45^(@)` and decreases afterwardsD. decreases up to `45^(@)` and increases afterwards

Answer» Correct Answer - C
54.

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal isA. `tan^(-1)((H)/(R ))`B. `tan^(-1)((2H)/(R ))`C. `tan^(-1) ((4H)/(R ))`D. `tan^(-1) ((4H)/(H))`

Answer» Correct Answer - C
Maximum height, `H = (u^(2) sin^(2) theta)/(2g) " "…(i)`
Horizontal range, `R = (u^(2) sin 2theta)/(g) = (2u^(2)sin theta costheta)/(g) " "...(ii)`
Divide (i) by (ii), we get
`(H)/(R ) = (tan theta)/(4) or tan theta = (4H)/(R ) or theta = tan^(-1) ((4H)/(R ))`
55.

The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is `20 m//s^(-1)`,its time of flight is `(g=10 m//s^(2))`A. `2s`B. `2sqrt2s`C. `1/2s`D. `1/sqrt2s`

Answer» Correct Answer - B
`P.E=K.ErArr1/2"mu"^(2) sin^(2)theta=1/2"mu"^(2) sin^(2)theta`
use`T=(2u sin theta)/g`
56.

Two particles A and B projected simultaneously from a point situated on a horizontal place. The particle A is projected vertically up with a velcity `v_(A)` while the particle B is projected up at an angle `30^(@)` with horizontal with velocity `v_(B)`. After 5s the particles were observed moving mutually perpendicular to each other. The velocity of projection of the particle `v_(A) and v_(B)` respectively are:A. `5ms^(-1),100 ma^(-1)`B. `100ms^(-1),50 ma^(-1)`C. `v_(A)` can have any value greater than `25ms^(-1),100ms^(-1)`D. `20ms^(-1),25 ms^(-1)`

Answer» Correct Answer - C
`T_(A)=(2U)/g=5s , Ugt(5g)/2Ugt25 m//s`.
For any value `U gt 25` the projectile will move only in vertical direction.So `B` should move in horizontal direction at maximum height it is horizontal.
`T_(a)=(2U sin theta)/g=5=U/(2xx10) theta=30^(@)`
`U=100m//s`
57.

Statement -1 : The graph between - Kinetic energy and vertical displacement is a stright line for a projectile. Statement -2 : The graph between kinetic energy and horizontal displacement is a straight is straight line for a projectile. Statement -3 : the graph between kinetic energy and time is a parabola for a projectile.A. F F FB. T T FC. T F TD. F F T

Answer» Correct Answer - 3
58.

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)-t` graph is a straight line with negative slope and positive interceptB. `x-t`graph is a straight line passing through originC. `y-t` graph is a straight line passing through originD. `u_(x)-t` graph is a straight line

Answer» Correct Answer - A::B::D
If `u` is the initial speed and `theta` the angle of projection.Then `v_(y)= u sin theta-"gt"` i.e., `x-t` graph is a straight line passing through the origin. `y=(u sin theta)t-1/2 "gt"^(2)`
i.e., `y-t`graph is parpbola i.e., `v_(x)-t` graph is a straight line parallel to `t`-axis.
59.

The displacement x of a particle in a straight line motion is given by `x=1-t-t^(2)`. The correct representation of the motion isA. B. C. D.

Answer» Correct Answer - B
`(dx)/(dt)=upsilon=1-2t`
Comparing with `upsilon=u+at`, we have, `u=-1ms^(-1)` and
`a=-2ms^(-2)`
At t = 0, x = 1m. Then, u and a both are negative. Hence, x-coordinate of particle will go on decreasing.
60.

The velocity `upsilon` of a particle as a function of its position (x) is expressed as `upsilon = sqrt(c_(1)-c_(2)x)`, where `c_(1)` and `c_(2)` are positive constants. The acceleration of the particle isA. `c_(2)`B. `-(c_(2))/(2)`C. `c_(1)-c_(2)`D. `(c_(1)+c_(2))/(2)`

Answer» Correct Answer - B
`upsilon^(2)=c_(1)-c_(2)x` comparing wiuth `upsilon^(2)=u^(2)+2as` we have
`a=-(c_(2))/(2)`
61.

The given graph shows the variation of velocity with displacement. Which one of the graphs given below correctly represents the variation of acceleration with displacement ? A. B. C. D.

Answer» Correct Answer - A
Given line has positive intercet but negative slopr. So its equation can be written as
`upsilon=-mx+upsilon_(0)["where ", m=tan theta=(upsilon_(0))/(x_(0))]` ….(i)
By differentiating with respect to time, we get
`(d upsilon)/(dt)=-m(dx)/(dt)=-m upsilon`
Now substituting the value of `upsilon` from Eq. (i), we get
`(d upsilon)/(dt)=-m[-mx+upsilon_(0)]=m^(2)x-m upsilon_(0)`
`therefore a=m^(2)x-m upsilon_(0)`
i.e., the graph between a and x should have positive slope but negsative intercept on a-axis. So, graph (a) is correct.
62.

`vecP,vecQ,vecR,vecS` are vector of equal magnitude. If `vecP+vecQ-vecR=0` angle between `vecP` and `vecQ` is `theta_(1)`.If `vecP+vecQ-vecS=0` angle between `vecP` and `vecS` is `theta_(2)`.The ratio of `theta_(1)` to`theta_(2)` isA. `1:2`B. `2:1`C. `1:1`D. `1:sqrt3`

Answer» Correct Answer - B
`vecP+vecQ=vecR,vecP+vecQ=vecS`,
`theta_(2)= "tan"^(-1)(Qsintheta_(1))/(P+Qcostheta_(1)),theta_(1)/theta_(2)=?`
63.

Resultant of two vectors of magnitude `P` and `Q` is of magnitude `Q`.If the magnitude of `vecQ` is doubled now the angle made by new resultant with `vecP` isA. `30^(@)`B. `90^(@)`C. `60^(@)`D. `120^(@)`

Answer» Correct Answer - B
`Q^(2)=P^(2)+Q^(2)+2PQcos theta` ,
`costheta=-P/(2Q),Tan alpha=(2Q sin theta)/(P+2Q cos theta)`
64.

The resultant of two vectors `vec(P)` and `vec(Q)` is `vec(R )`. If the magnitude of `vec(Q)` is doubled, the new resultant vector becomes perpendicular to `vec(P)`. Then, the magnitude of `vec(R )` is equal toA. `(P^(2)-Q^(2))/(2PQ)`B. `(P+Q)/(P-Q)`C. `Q`D. `P/Q`

Answer» Correct Answer - C
`R^(2)=P^(2)+Q^(2)+2PQ cos theta` when `theta`is doubled
`tan90=(2Q sin theta)/(P+2Q cos theta)`
65.

A boy is runing on the plane road with velocity v with a long hollow tube in his hand. The water is falling vertically downwards with velocity u. At water angle to the verticaly, he must inclined the tube the water drops enter it without touching its sides ?A. `tan^(-1)((v)/(u))`B. `sin^(-1)((v)/(u))`C. `tan^(-1)((u)/(v))`D. `cos^(-1)((v)/(u))`

Answer» Correct Answer - A
`tan theta = (upsilon_(H))/(upsilon_(V))=(upsilon)/(u)` or `tan^(-1)((upsilon)/(u))`
66.

A man walking with a speed of 3 km/h finds the rain drops falling vertically downwards. When the man increases his speed to 6km/h he find that the rain drops are falling making an angle of `30^(@)` with the vertical . Find the speed of the rain drops ( in km/h)

Answer» Correct Answer - 6
67.

Two vectors inclined at an angle `theta` have magnitude `3 N` and `5 N` and their resultant is of magnitude `4 N`. The angle `theta` isA. `90^(@)`B. `cos^(-1)(4/5)`C. `cos^(-1)(3/5)`D. `cos^(-1)(-3/5)`

Answer» Correct Answer - D
`R^(2)=P^(2)+Q^(2)+2PQcostheta`
68.

Two vectors having magnitude 12 and 13 are inclined at and angle` 45^(@)` to each other.find their resultant vector.

Answer» Correct Answer - 23.09 at ` 21.55^(@)` with vector of magnitude 13.
69.

Rain is falling at the speed of ` 25sqrt3 m//s` vertically. The wind blows west to east at a speed of 25 m/s. find the velocity of rain as experienced by a person standing on the ground.

Answer» Correct Answer - 50 m/s at ` 60^(@)` below horizontal towards east.
70.

A bomb is rest at the summit of a cliff breaks into two equal fragments.One of the fragments attains a horizontal velocity of `20sqrt3 ms^(-1)`.The horizontal distance between the two fragments, when their displacement vectors is inclined at `60^(@)` relative to each other is `(g=10ms^(-2))`A. `40sqrt3`B. `80sqrt3`C. `120sqrt3`D. `480sqrt3`

Answer» Correct Answer - D
`x=(u_(1)u_(2))(2sqrt(u_(1)u_(2)))/g "cot" theta/2`
71.

A bomb at rest is exploded and the pieces are scattered in all directions with a maximum velocity of `20ms^(-1)`.Dangerous distance from that spot is `(g=10 m//s^(2))`A. `10 m`B. `20 m`C. `30 m`D. `40 m`

Answer» Correct Answer - D
`R=u^(2)/g`
72.

The coach throws a baseball to a player with an initial speed at `20ms^(-1)` at an angle of `45^(@)` with the horizontal.At the moment the ball is thrown, the player is `50m` from coach.The speed and the direction that the player has to run to catch the ball at the same height at which it was released in `ms^(-1)` isA. `5/sqrt2`away from coachB. `5/sqrt2` towards from coachC. `sqrt2/5`towards the coachD. `sqrt2/5`away from the coach

Answer» Correct Answer - B
`R_("ball")=(u^(2)sin2theta)/g,s=vt`
73.

A man is 25 m behind a bus, when bus starts accelerating at `2 ms^(-2)` and man starts moving with constant velocity of `10 ms^(-1)`. Time taken by him to board the bus isA. 2 sB. 3 sC. 4 sD. 5 s

Answer» Correct Answer - D
Displacement of man = displacement of bus + 25
or `10t=(1)/(2)xx2xx t^(2)+25` or `10t=t^(2)+25`
Solving this equation, we get t = 5 s.
74.

A man can swim with a speed 4km/hr in still water. (a) How long does he takes to cross a river 1 km wide if the river flows steadily at 3 km/hr and makes his strokes normal to the river current? (b) How far down the river does he go when he go when he reaches the other bank?A. 500 mB. 600 mC. 750 mD. 850 m

Answer» Correct Answer - C
Time to cross the river,
`t=("Widht of river")/("Speed of man")=(1km)/(4kmh^(-1))=(1)/(4)h`
Distance moved along the river in time t,
`s=v_(r)xxt=3kmh^(-1)xx(1)/(4)h=(3)/(4)km=750m`
75.

Pick out the only vector quantity in the following list : temperature, pressure, impulse, time, power. Total path-length, energy. Gravitational potential, coefficient of friction, charge,A. Impulse, pressure and areaB. Impulse an areaC. Area and gravitational potentialD. Impulse and pressure

Answer» Correct Answer - B
We know that impulse `J=F.Deltat=Deltap`,where `F` is force, `Deltat` is time duration and `Deltap` is change in mementum.As `Deltap` is a vector quantity, hence impulse is also a vector quantity.Sometimes are can also be treated as vector.
76.

Consider the quantities , pressure, power, energy impulse, gravitational potential, electrical charge , temperature, area,Out of these, the only vector quantities are .A. impulse, pressure and areaB. impulse and areaC. area and gravitational potentialD. impulse and pressure

Answer» Correct Answer - B
We know that impulse `J=F. Deltat=Deltap`, where F is force, `Deltat` is time duration and `Delta p` is change in momentum. AS `Deltap` is a vector quantity , hence impulse is also a vector quantity. Sometimes area can also be treated as vector.
77.

A cricket fielder can throw the cricket ball with a speed ` v_0`. If he throws the ball while running with speed (u) at angle ` theta` to the horizontal, find (b) what will be time of flight ? (c ) what is the distance (horizontal range) form the point of projection at which the ball will land ? (d) find ` theta` at which he should throw the ball that would maxmise the horizontal range range as found in (c ). (e) how does ` theta` for maximum range change if ` u gt v_0, u=v_0, ult v_0` ? (f) how does ` theta` in (e) compare with that for ` u=0 (i.e., 45^@) ?A. `theta_(max)=cos^(-1)[(-u+sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]`B. `theta_(max)=cos^(-1)[(u+sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]`C. `theta_(max)=cos^(-1)[(-u+sqrt(u^(2)-8v_(0)^(2)))/(4v_(0))]`D. `theta_(max)=cos^(-1)[(-u+sqrt(u^(2)-4v_(0)^(2)))/(4v_(0))]`

Answer» Correct Answer - A
For horiontal range to be maximum,`(dR)/(dtheta)=0`
`rArr theta_(max)=cos^(-1)[(-u+sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]`
78.

A boy can throw a stone up to a maximum height of `10 m`. The maximum horizontal distance that the boy can throw the same stone up to will be :A. `20sqrt(2) m`B. `10m`C. `10 sqrt(2) m`D. `20m`

Answer» Correct Answer - D
`H_(max) = (u^(2))/(2 g) rArr 10 = (u^(2))/(2 g) rArr (u^(2))/(g) = 20`
`R_(max) = (u^(2))/(2 g) = 20 m`
79.

Rain is falling vertically with a speed of `35 m s^(-1)` . A woman rides a bicycle with a speed of ` 12 m s^(-1)` in east to west direction. In which direction should she hold her umbrella ?

Answer» Correct Answer - `10^(@)` with the vertical towards the west
80.

The width of a river is `2sqrt3km`.A boat is rowed in direction perpendicular to the banks of river.If the drift of the boat due to flow is `2 km`,the displacement of the boat is.A. `3 km`B. `6 km`C. `5 km`D. `4 km`

Answer» Correct Answer - D
`s=sqrt(d^(2)+x^(2))`
81.

A boat is moving towards east velocity 4 m/s w.r.t still water and river is flowing towards north with velocity 2 m/s and the wind blowing towards north with velocity 6 m/s. the direction of the flag blown over by the wind hpisted on the boat isA. North - westB. south- westC. `tan^(-1) (1/2)` with eastD. North

Answer» Correct Answer - 1
82.

A river of width a with straight parallel banks flows due north with speed u. The points O and A are on opposite banks and A is due east of O. Coordinate axes `O_x` and `O_y` are taken in the east and north directions respectively. A boat, whose speed is v relative to water, starts from O and crosses the river. If the boat is steered due east and u varies with `x as : u = x(a-x) v/a^2.` Find (a) equation of trajectory of the boat, (b) time taken to cross the river, (c) absolute velocity of boatman when he reaches the opposite bank, (d) the displacement of boatman when he reaches the opposite bank from the initial position.A. westB. southC. eastD. north

Answer» Correct Answer - C
When the boatman reaches the oppostie side, `x=a` or `v_(y)=0` (from eqution 1)
83.

A river of width a with straight parallel banks flows due north with speed u. The points O and A are on opposite banks and A is due east of O. Coordinate axes `O_x` and `O_y` are taken in the east and north directions respectively. A boat, whose speed is v relative to water, starts from O and crosses the river. If the boat is steered due east and u varies with `x as : u = x(a-x) v/a^2.` Find (a) equation of trajectory of the boat, (b) time taken to cross the river, (c) absolute velocity of boatman when he reaches the opposite bank, (d) the displacement of boatman when he reaches the opposite bank from the initial position.A. `a/v`B. `v/a`C. `(2a)/v`D. `(2v)/a`

Answer» Correct Answer - A
Time taken to cross the river is `t=a/v_(x)=a/v`
84.

An aeroplane flies 400 m north and 300m south and then flies 1200 m upwards then net displacement isA. 1200 mB. 1300 mC. 1400 mD. 1500 m

Answer» Correct Answer - A
An aeroplane flies 400 m north and 300 m south so the net displacement is 100 m towards north.
Then it flies 1200 m upwards so, `r=sqrt((100)^(2)+(1200)^(2))`
`=1204m ~= 1200 m`
The option should be 1204 m, because this value mislead one into thinking that net displacement is in upwards direction only.
85.

Which one of the following statements is true?A. A scalar quantity is the one that is conserved in a process.B. A scalar quantity is the one that can never take negative values.C. A scalar quantity is the one that does not vary from one point to antother in space.D. A scalar quantity has the same value for observers with different orientations of the axes.

Answer» Correct Answer - D
The statement is false. It is because kinetic energy (scalar quantity) is not conserved during inelastic collision.
(b) The statement is false. It is because the temperature (scalar quantity) can be negative.
(c ) The statement is false. It is because gravitational potential energy (scalar quantitiy) vary from point to point is space.
(d) The statement is true because the value of scalar does not change with orientation of axes.
86.

An aeroplane is moving with a velocity u. It drops a packet from a height h. The time t taken by the packet in reaching the ground will beA. `sqrt(((2g)/(h)))`B. `sqrt(((2u)/(g)))`C. `sqrt(((h)/(2g)))`D. `sqrt(((2h)/(g)))`

Answer» Correct Answer - D
The initial velocity of airoplane is horizontal. Hence, the vertical component of velocity of packet will be zero.
`therefore h=(1)/(2)g t^(2)`
`rArr t=sqrt((2h)/(g))`
87.

Two cliffs of heights `120 m` and `100.4 m` are separated bya horizontal distance of `16 m` if a car has to reach from the first cliff to the second the horizontal velocity of car should beA. `16 m//s`B. `4 m//s`C. `2 m//s`D. `8 m//s`

Answer» Correct Answer - D
`R=(usqrt(2(h_(1)-h_(2))))/g`
88.

Two balls are dropped from heights `h` and `2h` respectively from the earth surface. The ratio of time of these balls to reach the earth is.A. `1 : sqrt(2)`B. `sqrt(2) : 1`C. `2 : 1`D. `1 : 2`

Answer» Correct Answer - A
`t=sqrt((2h)/(g))` or `t prop sqrt(h) rArr (t_(1))/(t_(2))= sqrt((h)/(2h))=1: sqrt(2)`
89.

The displacement of a particle starting from rest (at `t = 0)` is given by `s = 6t^(2) - t^(3)`. The time in seconds at which the particle will attain zero velocity again, isA. 2B. 4C. 6D. 8

Answer» Correct Answer - B
Displacement of the particle
`s = 6t^(2)-t^(3)`
Velocity of the particle
`upsilon=(ds)/(dt)=(d)/(dt)(6t^(2)-t^(3))`
`upsilon=12t-3t^(2)`
For `upsilon = 0 rArr 12t=3t^() rArr t=4s`
90.

A particle is moving in a straight line. Its displacement at any instant t is given by `x = 10 t+ 15 t^(3)`, where x is in meters and t is in seconds. Find (i) the average acceleration in the intervasl t = 0 to t = 2s and (ii) instantaneous acceleration at t = 2 s.

Answer» The given equation, `x = 10 t + 15 t^(3)` and the variables are (i) t = 0 to t = 2 s (ii) t = 2 s
We need to clarify all the variables.
Velocity of particle, `v=(dx)/(dt)`
While calculating instantaneous velocity , we need to differentiate. Differentiating x w.r.t. time.
`v=(d)/(dt)(10 t+15t^(3))=10+45 t^(2)`
At `t = 0, V_(0) = 10+45(0)=10 ms^(-1)`
At t= 2 s,
`V_(2)=10+45xx(2)^(2)=10+180=190 ms^(-1)`
`Delta V= V_(2)-V_(1)=190-10=180 ms^(-1)`
`Delta t=2-0=2s`
`therefore a_(av)=(Delta V)/(Delta t)=(180)/(2)=90 ms^(-2)`
Now, differentiate, v with respect to time ti final the acceleration of the particle.
`a=(d)/(dt)(10+45 t^(2))=90 t`
At `t=2s, a = 90xx2 = 180 ms^(-2)`
The instantaneous acceleration of a particle at 2 is `180 ms^(-2)`.
91.

A body freely falling from the rest has velocity `v` after it falls through a height `h` the distance it has to fall down for its velocity to become double isA. 8 hB. 6 hC. 4 hD. 5 h

Answer» Correct Answer - C
`upsilon=sqrt(2gh)` or `upsilon prop sqrt(h)`
`rArr (upsilon_(1))/(upsilon_(2))=sqrt((h_(1))/(h_(2)))rArr ((upsilon_(1))/(upsilon_(2)))^(2)=(h)/(h_(2)) " " {because h_(1)=h}`
`rArr ((upsilon)/(2upsilon))^(2)=(h)/(h_(2)) " " {{:(because upsilon_(1)=upsilon),(therefore upsilon_(2)=2upsilon):}}`
`rArr h_(2)=4h`
92.

A ball is thrown horizontally from a cliff such that it strikes the ground after `5s`.The line of sight makes an angle `37^(@)` with the horizontal.The initial velocity of projection in `ms^(-1)` isA. 50B. `100/sqrt3`C. `100/sqrt2`D. `100/3`

Answer» Correct Answer - D
`Tantheta=h/R,T=usqrt((2h)/g)`and `R=uT`
93.

When air resistance is taken into account while dealing with the motion of the projectile which of the following properties of the projectile, shows an increases?A. rangeB. maximum heightC. speed at which it strikes the groundD. the angle at which the projectile strikes the ground.

Answer» Correct Answer - D
In the presence of air resistance, the range, maximum height, speed at which the projectile strikes the ground will decrease whereas the angle at which the projectile strikes the ground will increase.
94.

An object is projected with a velocity of `10 m//s` at an angle `45^(@)` with horizontal. The equation of trajectory followed by the projectile is `y = a x - beta x^(2)`, the ratio `alpha//beta` isA. `5`B. `10`C. `15`D. `20`

Answer» Correct Answer - B
`y = alpha x - beta x^(2) = x tan theta - (g x^(2))/(2u^(2) cos^(2) theta)`
`alpha = tan theta, beta = (g)/(2u^(2) cos^(2) theta)`
`(alpha)/(beta) = (tan theta)/(g//2u^(2) cos^(2) theta)`
`= (2u^(2) cos theta sin theta)/(g) = (2u^(2) cos^(2) theta tan theta)/(g)`
`= ((10)^(2))/(10)sin 90^(@) = 10`
95.

A heavy particle is projected with a velocity at an angle with the horizontal into the uniform gravitational field. The slope of the trajectory of the particle varies asA. B. C. D.

Answer» Correct Answer - A
`V_(x)=U cos theta , V_(y)=U sin theta-"gt"`
`Tan alpha=(U sin theta- "gt")/(U cos theta),Tan alpha=(sin theta)/(cos theta)-(g/(U cos theta))t`
Comparing with `y=c-mx`
96.

A ball P is dropped vertically and another ball Q is thrown horizontally with the same velocities from the same height and at the same time. If air resistance is neglected, thenA. ball P reaches the ground firstB. ball Q reaches the ground firstC. Both reach the ground at the same timeD. the respective masses of the two balls will decide the time

Answer» Correct Answer - C
Verical component of velocities of both the balls are same and equal to zero.
So, `t = sqrt((2h)/(g))`.
97.

A ball is thrown vertically upward from the 12 m level with an initial velocity of `18 m//s.` At the same instant an open platform elevator passes the 5 m level, moving upward with a constant velocity of `2 m//s.` Determine (`g = 9.8 m//s^2` ) (a) when and where the ball will meet the elevator, (b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator.A. `10.2m 9.8m//s`B. `12.3m 19.8m//s`C. `12m 10.2m//s`D. `12.5m 22m//s`

Answer» Correct Answer - B
a) Let the two meet at a distance `s` from ground.
Then `s-12=18t-1/2xx9.8xxt^(2)`..(i)
and `s-5=2t`…(ii) Solving these two equations, we get `t=3.65s` and `s=12.30m`
b) `v_(b)=18-(9.8)(3.65)=-17.8m//s` i.e, velocity of ball is `17.8 m//s` (downward) at the time of impact or relative velocity `=19.8 m//s` (downward)
98.

At a metro station, a girl walks up a stationary escalator in time `t_1` If she remains stationary on the escalator, then the escalator take her up in time `t_2`. The time taken by her to walk up the moving escalator will be.A. `((t_(1)+t_(2)))/(2)`B. `(t_(1)t_(2))/((t_(2)-t_(1)))`C. `(t_(1)t_(2))/((t_(2)+t_(1)))`D. `t_(1)-t_(2)`

Answer» Correct Answer - C
In this question, we have to find net velocity with respect to the Earth that will be equal to velocity of the girl plus velocity of escalator.
Let displacement is L, then
velocity of girl `upsilon_(g)=(L)/(t_(1))`
velocity of scalator `upsilon_(e )=(L)/(t_(2))`
Net velocity of the girl `= upsilon_(g)+upsilon_(e )=(L)/(t_(1))+(L)/(t_(2))`
If t is total time taken in covering distance L, then
`(L)/(t)=(L)/(t_(1)+(L)/(t_(2))`
`rArr t=(t_(1)t_(2))/(t_(1)+t_(2))`
99.

A ball is thrown straight upward with a speed v from a point h meter above the ground. The time taken for the ball to strike the ground isA. `(upsilon)/(g) sqrt(1-(2hg)/(upsilon^(2)))`B. `(upsilon)/(g)sqrt(1+(2hg)/(upsilon^(2)))`C. `sqrt(1+(2hg)/(upsilon^(2)))`D. `(upsilon)/(g)[1+sqrt(1+(2hg)/(upsilon^(2)))]`

Answer» Correct Answer - D
`-h=upsilon t-(1)/(2)g t^(2)` or `g t^(2)-2upsilon t-2h=0`
`therefore t=(2upsilon+sqrt(4upsilon^(2)+8gh))/(2g)=(upsilon)/(g)[1+sqrt(1+(2hg)/(upsilon^(2)))]`
100.

For body thrown horizontally from the top of a tower,A. the time of flight depends both on `h` and `v`B. the horizontal Range depends only on `v` but not on `h`C. the time of flight and horizontal Range depend on `h` but not on `v`D. The horizontal Range depends on both `v` and `h`

Answer» Correct Answer - D