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401.

The range of a projectile fired at an angle of `15^@` is 50 m. If it is fired with the same speed at an angle of `45^@` its range will beA. 60 mB. 71 mC. 100 mD. 141 m

Answer» Correct Answer - C
Horizontal range, `R = (u^(2) sin 2theta)/(g)`
For the same speed.
`R prop sin 2theta`
`therefore (R_(1))/(R_(2)) = (sin2 xx15^(@))/(sin2 xx 45^(@)) = (sin 30^(@))/(sin 90^(@))`
or `R_(2) = R_(1) (sin 90^(@))/(sin 30^(@)) = 50 m xx(1)/(((1)/(2))) = 100m`
Discussion
402.

A ball is dropped vertically from `a` height `d` above the ground . It hits the ground and bounces up vertically to a height ` (d)//(2). Neglecting subsequent motion and air resistance , its velocity `v` varies with the height `h` above the ground asA. B. C. D.

Answer» Correct Answer - A
For the given condition, initial height h = d and velocity of the ball is zero. When the ball moves downward, its velocity increases and it will be maximum, when the ball hits the ground and just after the collision, it becomes half hits and in opposite direction. As the ball moves upwards, its velocity again decreases and becomes zero at height d/2. This explanation match with graph (a).
Discussion
403.

Find the sum of the vectors ` 10hatj + 6hatj and 4hatj - 2hatj`

Answer» Correct Answer - `14hati + 4hatj` , magnitude = 14.56 , at an angle of ` 15. 94^(@)` with x-axis .
Discussion
404.

An object has a velocity, `v = (2hati + 4hatj) ms^(-1)` at time `t = 0s`. It undergoes a constant acceleration `a = (hati - 3hatj)ms^(-2)` for 4s. Then (i) Find the coordinates of the object if it is at origin at `t = 0` (ii) Find the magnitude of its velocity at the end of 4s.

Answer» (i) Here original position of the object.
` vecr_(0) = x_(0)hati + y_(0)hatj = 0 hati + 0 hatj`
Initial veclocity `vecv_(0) = v_(0)x hati + v_(0y) hatj = 2hatj + 4 hatj`
`veca = a_(x)hati + a_(y)hatj = hatj - 3hatj`
And t = 4 s.
Let the final co-rodinates of the object be (x,y) . then according to the equation (iii) derived in previous section.
` x = x_(0) + v_(0x)t + 1/2a_(x)t^(2)`
` = 0 +2 xx 4 + 1/2 (1) xx 4^(2)`
x = 16
`and y = y_(0) +v_(0y)t + 1/2 a_(y)t^(2) = 0 + 4 xx 4 + 1/2(-3) xx 4^(2)`
y = -8
Therefore the object lies. at (16 -8) at t= 4s.
(ii) Using eqution.
`vecv = vecv_(0) + vecat`
` Rightarrow vecv = (2hati + 4hatj) + ( hati - 3hatj) xx 4`
` = (2hatj + 4hatj) + ( 4hatj -12 hatj)`
` = (2+4 ) hatj + ( 4 -12)hatj`
` Rightarrow vecv = 6hati - 8o hatj ` : velocity at the end of 4s.
` |vecv| = sqrt(6^(2) +8^(2)) = 10 ms ^(-1)`
Its direcationn with x-axis, ` theta = tan^(-1) (-8)/6 = 53.13^(@)`
Discussion
405.

Three forces given by vectors ` 2hati + 2hatj. 2hati -2hatj and - 4hatj` are acting together on a point object at rest. The object moves along the direction.A. x -axisB. y-axisC. z-axisD. object does not move

Answer» Correct Answer - D
Discussion
406.

If `vecu=ahati+bhatj+chatk` with `hati,hatj,hatk`are in east, north and vertical directions, horizontal component of velocity of projectile isA. `a`B. `b`C. `sqrt(a^(2)+b^(2))`D. `sqrt(b^(2)+c^(2))`

Answer» Correct Answer - C
East and north are taken as ground.
Discussion
407.

Figure shows the orientation of two vectors `u` and `v`in the `XY` plane. if`vecu=ahati+bhatj` and `vecv=p hati+qhatj` which of the following is correct? A. `a` and `p` are positive while `b` and `q` are negativeB. `a,p` and `b` are positive while `q` is negativeC. `a,q` and `b` are positive while `p` is negativeD. `a,b,q` and `q` are all positive

Answer» Correct Answer - B
clearly from the diagram `u=ahati+bhatj`
As `u` is in the first quadrant, hence both components `a` and `b` will be positive For `v=phati+qhatj`,as it is in positive `x`-direction and located downward hence `x`-componenet `p` will be positive and `y`-component `q` will be negative.
Discussion
408.

A cyclist starts form centre ` O` of a circular park of radius ` 1 km` and moves along the path ` OPRQO` as shown Fig. 2 (EP).15. If he maintains constant speed of ` 10 ms^(-1)`, what is his acceleration at point (R )in magnitude and direction ? .A. 10 m `s^(-2)`B. 0.1 m `s^(-2)`C. 0.01 m `s^(-2)`D. 1 m `s^(-2)`

Answer» Correct Answer - B
The acceleration at point R is towards the centre of the circle.
Here, `v=10ms^(-1),r=1km=1000m`
Centripetal acceleration `a_(c)=(v^(2))/(r)=((10ms^(-1))^(2))/(1000m)=0.1ms^(-2)`
Discussion
409.

Figure 2 (EP). 13 shows the orientation of two vectors ` vec u` and `vec v` in the (XY) plane. If ` vec u = a hat i + b hat j` and ` vec v = p hat i+ q hat j` which of the following is correct ? .A. a and p are positive while b and q are negativeB. a,p and b are positive while q is negativeC. a, q and b are positive while p is negativeD. a, b, p and q are all positive

Answer» Correct Answer - B
Clearly from the diagram, `u=a hat(i) + b hat(j)`
As u is in the first quadrant, hence both components a and b will be positive.
For `v=p hat(i) + q hat (j)`, as it is in positive x-direction and located downward hence x-component p will be positive and y-component q will be negative.
Discussion
410.

The position of a particle is given by ` vec r= 3.0 t hat i - 2.0 t^2 hat j + 4.0 hat k m`, where (t) in seconds and the coefficients have the proper units for `vec r` to be in metres. (a) Find the ` vec v` and ` vec a` of the particle ? (b) What is the magnitude and direction of velocity of the particle at ` t= 2 s`?

Answer» `varr(t) = (3.0that(i)-2.0t^(2)hat(j)+4.0hatk))`m
a) `therefore` `vecv(t) = vec(dr)/(dt)= (3.0hat(i)-4.0hat(j))m//s`
and `veca(t) = (vec(dv)/(dt)= (-4.0hatj)m//s^(2)`
b) Magnitude of velocity at t=2.0 s,
`v_(t=2s) = sqrt((3.0)^(2)+(-4.0xx2)^(2))` = `sqrt(9+64)`= `sqrt(73)`.
`=8.54ms^(-1)`
This velocity will subtend an angle `beta` from x-axis, where `tanbeta= (-4.0xx2)/(3.0) = -2.667 = -2.6667`
`therefore beta=tan^(-1)(-2.6667)= -69.44^(@)= 69.44^(@)` from negative x-axis.
Discussion
411.

Two forces `F_(1)` and `F_(2)` are acting at a point, whose resultant is `F`.If `F_(2)` is doubled `F` is also doubled. If `F_(2)` is reversed then also `F` is doubled. Then `F_(1):F_(2):F` isA. `sqrt2:sqrt2:sqrt3`B. `sqrt3:sqrt3:sqrt2`C. `sqrt3:sqrt2:sqrt3`D. `sqrt2:sqrt3:sqrt2`

Answer» Correct Answer - D
`F^(2)=F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)costheta...(1)`
`4F^(2)=F_(1)^(2)+F_(2)^(2)+4F_(1)F_(2)costheta...(2)`
`4F^(2)=F_(1)^(2)+F_(2)^(2)-2F_(1)F_(2)costheta...(3)`
Discussion
412.

Let `r_(1)(t)=3t hat(i)+4t^(2)hat(j)` and `r_(2)(t)=4t^(2) hat(i)+3t^(2)hat(j)` represent the positions of particles 1 and 2, respectiely, as function of time t, `r_(1)(t)` and `r_(2)(t)` are in metre and t in second. The relative speed of the two particle at the instant t = 1s, will beA. 1 m/sB. `3sqrt(2) m//s`C. `5sqrt(2) m//s`D. `7sqrt(2)m//s`

Answer» Correct Answer - C
Given,
`r_(1)=3t hat(i)+4t^(2)hat(j)`
`therefore (dr_(1))/(dt)=3hat(i)+8t hat(j)`
At t = 1 s
`upsilon_(1)=(dr_(1))(dt)=3 hat(i)+8hat(j)`
Again, `r_(2)(t)=4t^(2)hat(i)+3hat(t)hat(j)`
`(dr_(2))/(dt)=8hat(t)hat(j)+3hat(j)`
At t = 1 s
`upsilon_(2)=(dr_(2))/(dt)=8hat(i)+3hat(j)`
Relative velocity `=upsilon_(1)-upsilon_(2)=-5hat(i)+5hat(j)`
`=sqrt((5)^(2)+(5)^(2))=5sqrt(2)ms^(-1)`
Discussion
413.

A body X is projected upwards with a velocity of `98 ms^(-1)`, after 4s, a second body Y is also projected upwards with the same initial velocity . Two bodies will meet afterA. 8 sB. 10 sC. 12 sD. 14 s

Answer» Correct Answer - C
Let t second be the time of flight of the first body after meeting, then (t - 4) second will be the time of flight of the second body.
Since, `h_(1)=h_(2)`
`therefore 98t-(1)/(2)g t^(2)=98(t-4)g(t-4)(2)`
On solving, t = 12 s
Discussion
414.

A particle is fired from `A` in the diagonal plane of a building of dimension `20m`(length) `xx` `15m` (breadth) `xx` `12.5 m`(height), just clears the roof diagonally & falls on the other side of the building at `B`.It is observed that the particle is travelling at an angle `45^(@)` with the horizontal when it clears the edges `P` and `Q` of the diagonal. Take `g=10m//s^(2)`. The range that is `AB` will be: A. `5sqrt10m`B. `25sqrt3m`C. `5sqrt15m`D. `25sqrt5m`

Answer» Correct Answer - B
Range `(U^(2) sin 2theta)/g`
Discussion
415.

A particle is fired from `A` in the diagonal plane of a building of dimension `20m`(length) `xx` `15m` (breadth) `xx` `12.5 m`(height), just clears the roof diagonally & falls on the other side of the building at `B`.It is observed that the particle is travelling at an angle `45^(@)` with the horizontal when it clears the edges `P` and `Q` of the diagonal. Take `g=10m//s^(2)`. The speed of projection of the particle at `A` will be: A. `5sqrt10m//s`B. `10sqrt5m//s`C. `5sqrt15m//s`D. `5sqrt5m//s`

Answer» Correct Answer - B
`V_(y)^(2)-U_(y)^(2)=2a_(y)S_(y)`
`125-(U sin theta)^(2)=-2(10)125 rarr (1)`
`5sqrt10 cos 45=V cos theta rarr(2)`
On solving `U=sqrt500m//s,theta=Tan^(-1)sqrt3=60^(@)`
Discussion