DIRECT ARYLATION OF HETEROCYCLES

and Average speed =

Total Distance covered Total time taken

(A) statement-1 is true, statement-2 is true; statement-2 is a correct explanation for statement-1.

(B) statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1.

(C) statement-1 is true, statement-2 is False.

(D) statement-1 is False, statement-2 is True.

Sol. Let the direction of initial velocity 'u' and constant acceleration 'a' of the particle be opposite. Then after the particle turns back and acquire a velocity of magnitude larger than 3u, the magnitude of average velocity

u V 2

 shall be greater than v. Hence statement-1 is false.

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142. A block of mass m starts from rest at height h on a frictionless inclined plane. The block slides down the plane, travels a distance d on a rough horizontal surface (before coming to rest for the first time) with co-efficient of kinetic friction µ & compresses a spring with force constant k through a distance x before momentarily coming to rest. Then the block travels back across the rough surface, sliding up the plane. The correct expression for the maximum height h' which the block reaches on its return is:

(A) h' = h – 2µ d (B) h' = h + 2µ d

(C) h' = h + 2 µ d +

k

²

mg x

(D) h' = h – 2 µ d –

k

²

mg x

Sol. W_G + W_SF + W_FF +  K.E.

mg (h – h') + 0 + µ mg (– 2d) = 0 h – h' – 2 µd = 0

h' = h – 2µd

 (A)

143. A particle is projected vertically upwards with a speed of 16 m/s, after some time, when it again passes resistance is same during upward and

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downward motion. Then the maximum height attained by the particle is (Take g = 10 m/s²):

(A) 8 m (B) 4.8 m

(C) 17.6 m (D) 12.8 m

Sol. (A)

From work energy theorem for upward motion

1

2

^{m (16)2} = mgh + W (work by air resistance) for downward motion

 

²

1 m 8 mgh W

2

 

   

^{2 2}

1 16 8 2gh

2

    or h = 8 m.

144. A horse drinks water from a cubical container of side 1 m. The level of the stomach of horse is at 2 m from the ground. Assume that all the water drunk by the horse is at a level of 2 m from the ground. Then minimum work done by the horse in drinking the entire water of the container is (take _water = 1 000 kg/m ³ = 10 m/s²):

(A) 10 kJ (B) 15 kJ

(C) 20 kJ (D) zero

Sol. The mass of water is

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= 1 × 10³ kg

 The increase is potential energy of water is

= 1 × 10³ × 10 × 1.5

= 15000 J

= 15 kJ

145. A man places a chain (of mass 'm' and length 'l') on a table slowly.

Initially the lower end of the chain just touch the table. The man drops the chain when half of the chain is in vertical position. Then work done by the man is this process is:

(A)

mg



2

(B)

mg



4

(C)

3mg



8

(D)

mg



8

Sol. (C)

The work done by man in negative of magnitude of decrease in potential energy of chain.

U mg m g 2 2 4

  

=

3mg 8 W 3mg

  

8

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146. A ball is suspended from the top of a cart by a string of length 1.0 m.

The cart and the ball are initially moving to the right at constant speed V, The cart comes to rest after colliding and sticking to a fixed bumper, as in figure II. The suspended ball swings through a maximum angle 60°. The initial speed V is (take g = 10 m/s²)

(A)

10 m / s

(B)

2 5 m / s

(C)

5 2 m / s

(D) 4 m/s

Sol. As string does no work on the ball, energy conservation can be applied

1 mV

2

2

= mg (L – V = ^{2gL 1 cos}



^{ }



on putting values V =

10 m / s

147. S₁: Path of a particle moving along a straight line with respect to the observer moving along another straight line must be straight line.

S₂: A man is standing inside a lift which is moving upwards. He shall fell more weight as compared to when lift was are rest.

S₃: A block moves down the inclined surface of a wedge. The wedge lies on smooth horizontal surface. The work done by normal reaction (exerted by wedge on the block) on the block must be zero.

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S₄: When ever a block is placed on top of the rough horizontal surface of another block and lower block is moved horizontally, work done by friction on block is always positive.

(A) F F T T (B) F F F T

(C) F T F T (D) T T F T

Sol. If the initial relative velocity and relative acceleration are neither parallel nor anti parallel, the particle shall move in any not straight line curve.

Hence S₁ is false.

If the lift is retarding upwards, he shall feel lighter. Hence S₂ is false.

For moving wedge, the work done by normal reaction on block (exerted by wedge) is non zero. Hence S₃ is false.

The friction on 'upper block is in the direction of its motion.

w.d. is positive. Hence S₄ is true.

148. A light, inextensible string attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the string is pulled over a pulley, of negligible mass and friction and fixed at height h = 3m from the ground level. The cart slides from X₁ = 3 3m to x₂ = 4m and during the move, tension in the string is kept constant 50 N. Find change in kinetic energy of the cart in joules (Use 3 = 1.7)

Sol. Displacement of the point of 'A of the string

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=

 

^{3 3 2 }

^{ }

^{32  42 32}

= 6 – 5 = 1m

Change in kinetic energy = work done by tension

= 50 × 1 = 50 Joule.

149. Two identical blocks A and B are placed on two inclined planes neglect air resistance and other friction.

Read the following statements and choose the correct options.

Statement-I:

Kinetic energy of 'A' on sliding to J will be greater than the kinetic energy of B on falling to M.

Statement-II:

Acceleration of 'A' will be greater than acceleration of 'B' when both are released to slide on inclined plane.

Statement-III:

Work done by external agent to move block slowly from position B to O is negative.

(A) only statement-I is true (B) only statement-II is true (C) only I and III are true (D) only II and III are true

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Sol. Statement-I:

Work done by gravity is same for motion from A to J and B to M for equal mass. So K.E. with be equal.

Statement-II:

Acceleration = g sin  sin _A > sin _B

h h



2

Statement-III:

W_g + W_ext = 0

(Because moved slowly) W_ext = – Wg

From B to O :

W_g is positive so W_ert < 0.

150. Block A is released from rest when the extension in the spring is x₀ (x₀ <

mg/k). The maximum downwards displacement of the block is (there is no friction):

(A)

2Mg

₀

K



2x

(B)

Mg x

₀

2K



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(C)

2Mg x

₀

Determine the time t from the beginning of motion to the moment of impact, neglecting friction losses.

Sol. Since the wall is smooth, the impact against the wall does alter the vertical component of the ball velocity. Therefore, the total time t1 of motion of the ball is the total time of the ascent and descent of the body thrown upwards at a velocity v₀ gravitational field.

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Consequently, t₁ = 2v₀ sin

. g

 The motion of the ball along the

horizontal is the sum of two motions. Before the collision with the wall, it moves at a velocity v₀ cos . After the collision, it traverses the same distance backwards, but at a different velocity. In order to calculate the velocity of the backward motion of the ball, it should be noted that the velocity at which the ball approaches the wall (along the horizontal) is v₀ cos  + v. Since the impact is perfectly elastic, the ball moves away from the wall after the collision at a velocity v₀ cos  + v. Therefore, the ball has the following horizontal velocity relative to the ground.

(v₀ cos  + v) + v = v₀ cos  + 2v.

If the time of motion before the impact is t, by equating the distances covered by the ball before and after the collision, we obtain the

Q.152. A small ball move at a constant velocity v along a horizontal surface and at point A falls into a vertical well of depth H and radius r. The velocity v of the ball forms an angle  with the diameter of the well

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drawn through point A (See Figure). Determine the relation between v, H, r, and  for which the ball can "get out" of the well after elastic impacts with the walls. Friction losses should be neglected.

Sol. The top view of the trajectory of the ball. Since the collisions of the ball with the wall and the bottom of the well are elastic, the magnitude of the horizontal component of the ball velocity remains unchanged and equal to v. The horizontal distances between points of two successive collisions are AA₁ = A₁A₂ = A₁A₂ = ... = 2r cos . The time between two successive collisions of the ball with the wall of the well is t₁ = 2r cos

.

v



The vertical component of the ball velocity does not change upon a collision with the wall and reverses its sign upon a collision with the bottom. The magnitude of the vertical velocity component for the first impact against the bottom is 2gH, and the time of motion from the top to the bottom of the well is t₂ = 2H

g .

The vertical plane development of the polyhedron A₁A₂A₃... On this development, the segments of the trajectory of the ball inside the well are parabolas (complete parabolas are the segments of the trajectory, between successive impacts against the bottom). The ball can "get out"

of the well if the moment of the maximum ascent along the parabola

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coincides with the moment of an impact against the wall (i.e. at the moment of maximum ascent, the ball is at point A_n of the well edge).

The time t₁ is connected to the time t₂ through the following relations:

n₁ = 2kt₂, where n and k are integral and mutually prime numbers.

Substituting the values of t₁ and t₂, we obtain the relation between v₁ H, r, and  for which the ball can "get out" of the well:

nr cos k 2H.

v g

 

Q.153. A cannon fires from under a shelter inclined at an angle  to the horizontal. The cannon is at point A at a distance l from the base of the shelter (point B). The initial velocity of the shell is v₀, and its trajectory lies in the plane of the figure. Determine the maximum range L_max of the shell.

Sol. From all possible trajectories of the shell, we choose the one that touches the shelter. Let us analyze the motion of the shell in the coordinate system with the axes directed.

The "horizontal" component (along the axis A_x) of the initial velocity of the shell in this system is v_0x = v₀ cos ( – ), where  is the angle formed by the direction of the initial velocity of the shell and the horizontal. Point C at which the trajectory of the shell touches the shelter determines the maximum altitude h' of the shell above the horizontal. Figure shows that h'

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= l sin . The projection of the total velocity v of shell on the axis Ay is zero Hence it follows, in particular, that if

2

v0  2gl cos sin  gl sin2

By hypothesis, none of the shell trajectories will touch the shelter, and the maximum range L_max

=

4

be fired to touch the shelter will be

t

20

is satisfied by hypothesis, which means that the condition ₁ >

4

their drivers see the hailstones bounced by the windscreen of their cars in the vertical direction? Assume that hailstones fall vertically.

Sol. Let us suppose that hall falls along the vertical at a velocity v. In the reference frame fixed to the motorcar, the angle of incidence of hailstones on the windscreen is equal to the angle of reflection. The velocity of a hailstone before it strikes the windscreen is v – v₁

Since hailstones are bounced vertically upwards (from the viewpoint of the driver) after the reflection the angle of reflection, and hence the cot 2₁. Therefore, we obtain the following ratio of the velocities of the two motor cars.

Q.155. A car must be parked in a small gap between the cars parked in a row along the pavement. Should the car be driven out forwards or backwards for the manoeuvre if only its front wheels can be turned?

Sol. Let a car get in a small gap between two other cars. It is parked relative to the pavement.

Is it easier for the car to be driven out of the gap by forward or backward motion? Since only the front wheels can be turned, the centre O of the circle along which the car is driven out for any

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manoeuvre (forward or backward) always lies on the straight line passing through the centres of the rear wheels of the ear.

Consequently, the car being driven out is more likely to touch the hind car during backward motion then the front car during forward motion (the centre of the corresponding circle is shifted backwards relative to the middle of the car). Obviously, a driving out is a driving in inversed in time. Therefore, the car should be driven in a small gap by backward motion.

Q.156. An aeroplane flying along the horizontal at a velocity v₀ starts to ascend, describing a circle in the vertical plane. The velocity of the plane changes with height h above the initial level of motion according to the law v² = v²₀ 2a h.₀ The velocity of the plane at the upper point of the trajectory is v₁ =

v

⁰

2 .

Determine the acceleration a of the plane at the moment when its velocity is directed vertically upwards.

Sol. Let us consider the motion of the plane starting from the moment it goes over to the circular path. By hypothesis at the upper point B of the path, the velocity of the plane is v₁ =

v

⁰

,

2

and hence the radius r of the circle described by the plane can be found from the relation

2 2

0 0 0

v v 2a .2r, 4  

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Which is obtained from the law of motion of the plane for h = 2r. At point C of the path where the velocity of the plane is directed upwards, the total acceleration will be the sum of the centripetal acceleration a_C

=

2 2 2

C C 0 0

v (v v 2a r,

r   where v_C is the velocity of the plane at point C) and the tangential acceleration a_t (this acceleration is responsible for the change in the magnitude of the velocity).

In order to find the tangential acceleration, let us consider a small

Let us divide both sides of the obtained relation by the time interval t during which this change takes place.

2 2 when the velocity has the upward direction is then

a =

=

a 100

⁰

3 .

Q.157. A bobbin rolls without slipping over a horizontal surface so that the velocity v of the end of the thread (point A) is directed along the horizontal. A board hinged at point B leans against the bobbin. The inner and outer radii of the bobbin are r and R respectively.

Determine the angular velocity  of the board as a function of an angle .

Sol. Let the board touch the bobbin at point C at a certain instant of time.

The velocity of point C is the sum of the velocity v₀ of the axis O of bobbin and the velocity of point C (relative to point O), which is tangent to the circle at point C and equal in magnitude to v₀ (since Were is no slipping). If the angular velocity of the board at this instant is , the linear velocity of the point of the board touching the bobbin will be R tan^–1

Since there is no slipping of the bobbin over the horizontal surface, we can write

v

0

v R



R r .



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Therefore, we obtain the following expression for the angular velocity thickness is half that of the initial tape?

Sol. The area of the spool occupied by the wound thick tape is S₁ = case. Since the lengths of the tapes are equal and the tape thickness in the latter case is half that in the former case, we can write

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l =



^{2f i2}



2 2 2

Consequently, the final radius r'_f of the wound part in the latter case is r'_f = 5 r .₁

The numbers of turns N₁ and N₂ of of the spool for the former and latter winding can be written as

N₁ = ₁

 

¹ the winding radius be reduced by half again?

Sol. Let the initial winding radius be 4r. Then the decrease in the winding area as a result of the reduction of the radius by half (to 2r) will be S =  (10 r² – 4 r²) = 12 r²,

Which is equal to the product of the length l₁ of the wound tape and its thickness d. The velocity v of the tape is constant during the operation of the tape-recorder, hence l₁ = vt₂, and we can write

12r² = vt₁d …(i)

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When the winding radius of the tape on the cassette is reduced by half again (from 2r to r), the winding area is reduced by  (4 r² – r²) = 3 r², i.e.

3r² = vt₂d …(ii)

Where t2 is the time during which the winding radius will be reduced in the latter case. Dividing equation (i) and (ii) term wise, we obtain

t₂ =

t

¹

5 min.

4



Q.160. Two rings O and O' are put on two vertical stationary rods AB and A'B' respectively. An inextensible thread is fixed at point A' and on ring O and is passed through ring O' (See figure). Assuming that ring O' moves downwards at a constant velocity v₁, determine the velocity v₂ of ring O if AOO' = .

Sol. Let us go over to the reference frame fixed to ring O' c. In this system the velocity of ring O is

v

¹

cos

 and is directed upwards since the thread is inextensible and is pulled at a constant velocity v₁ relative to ring Oc. Therefore, the velocity of ring O relative to the straight line AA'c (which is stationary with respect to the ground) is

v₂ =

and is directed upwards. time, the load will move along the wedge over the same distance s, and its velocity relative to the wedge is v_rel = at and directed upwards along the wedge. The velocity of the load relative to the ground is v₁ = v_rel + v_wed, i.e.

v₁ = 2v_wed sin 2 sin t,

2 2

    

and the angle formed by the velocity v₁ with the horizontal is

2

wedge relative to the ground is

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a₁ 2a sin

. 2



Q.162. An ant runs from an ant-hill in a straight line so that its velocity is inversely proportional to the distance from the centre of the ant-hill.

When the ant is at point A at a distance l₁ – 1 m from the centre of the ant-hill its velocity v₁ = 2 cm/s. What time will it take the ant to run from point A to point B which is at a distance l₁ = 2 m from the centre of the ant-hill?

Sol. The velocity of the ant varies with time according to a nonlinear law.

Therefore, the mean velocity on different segments of the path will not be the same, and the well-known formulas for mean velocity cannot be used here. formula suggests the idea of the solution of the problem, we plot the dependence of

m

1

v (l) on 1 for the path between points A and B. The graph is segment of a straight line. The hatched area S under this segment is numerically equal to the sought time. Let us calculate this area.

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S = ^{1 2}



2 1



Thus, the ant reaches point B in the time

t =

Q.163. Three school boys, Sam, John, and Nick are on merry-go-rounds-Sam and John occupy diametrically opposite points on a merry-go-round of radius r. Nick is on another merry-go-round of radius R. The positions of the boys at the initial instant are shown in figure.

Considering that the merry-go-round touch each other and rotate in the same direction at the same angular velocity , determine the nature of motion of Nick from John's point of view and of Sam from Nick's point of view.

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Sol. Let the merry-go-round turn through a certain angle . We construct a point O (OA = R) such that points 0, S, A and J lie on the same straight line. Then it is clear that ON = R + r at any instant of time. Besides, point O is at rest relative to John (Sam is always opposite to John).

Therefore, from John's point of view, Nick translates in a circle of radius R + r with the centre at point O which moves relative to the ground in a circle of radius R with the centre at point A. From Nick's point of view, Sam translates in a circle of radius R + r with the centre at point O which is at rest relative to Nick. However, point O moves relative to the ground in a circle of radius r with the centre at point B.

Q.164. A hoop of radius r rests on a horizontal surface. A similar hoop moves past it at a velocity v. Determine the velocity VAof the upper point of

"intersection" of the hoops as a function of the distance d between their centres, assuming that the hoops are thin, and the second hoop is in contact with the first hoop as it moves past the latter.

"intersection" of the hoops as a function of the distance d between their centres, assuming that the hoops are thin, and the second hoop is in contact with the first hoop as it moves past the latter.

In document Impregnated Cobalt, Nickel, Copper and Palladium Oxides on Magnetite: Nanocatalysts (página 134-141)