# Physics 9702 Doubts | Help Page 80

__Question 422: [Kinematics > Linear motion]__
Stone is dropped from top of a tower
of height 40 m. Stone falls from rest and air resistance is negligible.

What time is taken for stone to fall
the last 10 m to the ground?

A 0.38 s B 1.4 s C
2.5 s D 2.9 s

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q8*

__Solution 422:__**Answer: A.**

Consider the equation for uniformly
accelerated motion: s = ut + ½ at

^{2}
First, let’s calculate the time t

_{1}for the stone to fall a distance of 40m to reach the ground.
s = ut + ½ at

^{2}
40 = 0 + 0.5(9.81)t

_{1}^{2}
Time t

_{1}= 2.86s.
Now, let’s calculate the time t

_{2}for the stone to fall a distance of 30m.
30 = 0 + 1.5(9.81)t

_{2}^{2}
Time t

_{2}= 2.47s
Thus, the additional time required
to fall the last 10m is given by

Time = t

_{1}– t_{2}= 0.38s

__Question 423: [Current of Electricity]__
Two wires P and Q made of same material
are connected to same electrical supply. P has twice the length of Q and
one-third of the diameter of Q, as shown.

What is ratio

**of**current in P**to**current in Q?
A 2 / 3 B 2 / 9 C 1 / 6 D 1 /
18

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q35*

__Solution 423:__**Answer: D.**

Ohm’s law: V = IR

Current I = V / R (the current is inversely proportional
to the resistance)

From Kirchhoff’s law, there is the
same p.d. across P and Q.

So, Ratio = I

_{P}/ I_{Q}= R_{Q}/ R_{P}
Resistance R of a wire = ρL/A = ρL/ (π[d/2]

^{2})
For wire Q, the length is L and the
diameter is 3d (instead of d in the above equation).

Resistance of wire Q, R

_{Q}= ρL/(π(1.5d)^{2}) = (4/9)x(ρL/πd^{2})
For wire P, the length is 2L (instead
of L in the equation for resistance R of a wire) and the diameter is d.

Resistance of wire P, R

_{P}= 2ρL/(π(0.5d)^{2}) = 8 x(ρL/πd^{2})
Ratio = I

_{P}/ I_{Q}= R_{Q}/ R_{P}= (4/9) / 8 = 1 / 18

__Question 424: [Current of Electricity > Internal resistance]__
Power supply of electromotive force
(e.m.f.) 12 V and internal resistance 2.0 Ω is connected in series with 13 Ω
resistor.

What is power dissipated in the 13 Ω
resistor?

A 8.3 W B 9.6 W C
10 W D 11 W

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q32*

__Solution 424:__**Answer: A.**

Power P dissipated in resistor = I

^{2}R
Current I in circuit = V / R

_{T}= 12 / (2.0+13) = 0.80A
Power P dissipated in resistor = I

^{2}R = (0.8)^{2}(13) = 8.3W

__Question 425: [Forces > Newton’s laws of motion]__
Box of mass 8.0 kg rests on
horizontal, rough surface. A string attached to box passes over smooth pulley
and supports a 2.0 kg mass at its other end.

When box is released, a friction
force of 6.0 N acts on it.

What is acceleration of the box?

A 1.4 m s

^{–2}B 1.7 m s^{–2}C 2.0 m s^{–2}D 2.5 m s^{–2}**Reference:**

*Past Exam Paper – November 2008 Paper 1 Q11 & June 2012 Paper 12 Q13 & June 2015 Paper 11 Q13*

__Solution 425:__**Answer: A.**

The weight of the 2.0kg mass acts
downwards. So, the 2.0kg mass pulls the 8.0kg box to the right with a force (equal
to its weight) of

Weight of 2.0kg mass = 2.0 (9.81) =
19.62N

The force of friction opposes motion
and thus it acts to the left.

Resultant force on the system =
19.62 – 6.0 = 13.62N = ma.

It is important to note that the
system consists of an 8.0kg box AND a 2.0kg mass. So, the total mass in the
system is 8.0 + 2.0 = 10.0kg. So, both the box and the mass need to be
accelerated.

ma = 13.62N

Acceleration a = 13.62 / 10 = 1.36 ≈
1.4ms

^{-2}
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