DEPARTMENT OF GENERAL STUDIES
YANBU INDUSTRIAL COLLEGE
Semester I, 1435 – 1436 (H) / 2014 – 2015 (G)
PHY– 101, (General Physics)
Final Examination, Monday, 5/1/ 2015
Time Allowed:
2.5 Hours
Student ID: ________________________ Max. Marks: 35
Please answer all questions in answer sheet.
Question (1)
(7 Marks)
(a) Give two examples of derived quantities and write the SI unit for each one of them.
(b) Write four equations of motion in one dimension with constant acceleration.
(c) A shotputter throws the shot with an initial speed of 18 m/s at 30 o angle to the horizontal.
Calculate the horizontal distance traveled by the shot if it leaves the athlete’s hand at a
height of 2.0 m above the ground.
Question (2)
(7 Marks)
(a) (i) State (in words) Pascal’s Principle. (ii) Give two examples of Pascal’s Principle.
(b) Define specific heat.
(c) A 13kg box is released on a 37o incline and accelerates
down the incline at 2 m/s2 as shown in the figure beside.
(i) Draw the free body diagram of the box.
(ii) Find the friction force impeding motion of the box.
(iii) What is the coefficient of kinetic friction?
Question (3)
(7 Marks) (a) (i) State (in words) Work-Energy principle.
(ii) Define electric current.
(b) How much work (in Joules) can a 4 hp motor do in 2 h.
(c) The roller-coaster car shown in the figure below is dragged up to point 1 where
it is released from rest. Assuming no friction, calculate the speed at point 2.
1
2
35m
1
20m
Question (4)
(a) Draw electric field lines:
(7 Marks)
(i) around a single negative charge.
(ii) between two parallel plates having opposite charges.
(b) If wind blows at 40 m/s over a house, what is the net force on the roof if its area is
220 m2 and is flat.
(c) When a 250 g piece of iron at 190 ℃ is placed in an 85 g aluminum calorimeter cup
containing 300 g of a certain liquid at 10 ℃, the final temperature is observed to be
40oC. Estimate the specific heat of the liquid.
Question (5)
(a) A steady current of 3 A exists in a wire for 5 min.
(7 Marks)
(i) How much total charge passed by a given point in the circuit during those 5 min?
(ii) How many electrons would this be?
(b) The power supply for a pulsed nitrogen laser has a 0.07 𝜇F capacitor with a maximum
voltage rating of 25 kV. Estimate how much energy could be stored in this capacitor.
(c) Three charged particles are arranged in a line as shown in the figure below. Find the
magnitude and direction of the net electrostatic force on charge (𝑸𝟑 = −𝟑𝝁C) due to
the other two charges.
𝟎. 𝟒𝒎
𝟎. 𝟐𝒎
𝑸𝟏 = −𝟏𝟏𝝁𝑪
𝑸𝟐 = +𝟒𝝁𝑪
𝑸𝟑 = −𝟑𝝁𝑪
Constants
1hp = 746 W
𝜌𝑎𝑖𝑟 = 1.3 kg/m3
2
g = 9.8 m/s
1𝑒 =1.6 ×10-19 C
1
𝑘=
= 9 × 109 N. m2 /C 2
4𝜋𝜀𝑜
GOOD LUCK
2
caluminum= 900 J/kg.℃
ciron= 450 J/kg.℃
DEPARTMENT OF GENERAL STUDIES
YANBU INDUSTRIAL COLLEGE
Semester I, 1435 – 1436 (H) / 2014 – 2015(G)
PHY– 101, (General Physics)
Key of Final Examination
* Please consider the following:
(1) If student made a mistake in one part which affects the results of other
parts, then a mark should be deducted from the first part only, and we
should award the student the marks for the other parts.
(2) If the student did not write the unit of (the final result only) correctly, then
he should not be awarded any mark for the final result (if the mark of the
final result is 0.5).
(3) Pay attention to the notes listed in the table of answer key.
3
Q# Part
Q1) (a) 1) speed,
2) acceleration,
Solution
SI unit is (m/s)
SI unit is (m/s2)
Mark
NOTES
0.5, 0.5 0.5 mark for
0.5, 0.5 each
quantity
and 0.5
mark for
each SI
unit.
OR
Any two derived quantities with their SI units.
(b)
𝑣 = 𝑣𝑜 + 𝑎 𝑡
OR
𝑣 2 = 𝑣𝑜2 + 2𝑎 (𝑥 − 𝑥𝑜 )
1
𝑥 = 𝑥𝑜 + 𝑣𝑜 𝑡 + 𝑎 𝑡 2
2
𝑣 + 𝑣𝑜
𝑣̅ =
(c)
2
𝑣𝑜 = 18 𝑚/𝑠, 𝜃0 = 30𝑜 ,
𝑎=
∆𝑣
∆𝑡
0.5
0.5
0.5
0.5
𝑦0 = 2𝑚
𝑣𝑥𝑜 = 𝑣𝑜 𝑐𝑜𝑠𝜃𝑜 = 18 cos 30 = 15.59 𝑚/𝑠
𝑣𝑦𝑜 = 𝑣𝑜 𝑠𝑖𝑛𝜃𝑜 = 18 sin 30 = 9 𝑚/𝑠
𝑥 = 𝑥𝑜 + 𝑣𝑥𝑜 𝑡
𝑡 =? ?
1
𝑦 = 𝑦𝑜 + 𝑣𝑦𝑜 𝑡 − 𝑔𝑡 2
2
1
0 = 2 + 9𝑡 − (9.8)𝑡 2
2
𝑡1 = −0.2𝑠, 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑
𝑡2 = 2.04 𝑠
𝒙 = 𝟎 + 𝟏𝟓. 𝟓𝟗 × 𝟐. 𝟎𝟒 = 𝟑𝟏. 𝟖𝒎
Total Marks
4
0.5
0.5
0.5
0.5
0.5
0.5
7
If student
writes 𝒈
instead of a,
give him
one mark
out of two.
Q# Part
Solution
(i)
Pascal’s
Principle:
If an external pressure is
Q2) (a)
applied to a confined fluid, the pressure at every
point within the fluid increases by that amount.
(ii) (a) Hydraulic lift,
(b) Hydraulic brakes in car.
OR
(b)
Any other two examples of Pascal’s Principle.
Specific heat: heat needed to raise 1 gram
(kg) of material by 1 ℃.
Mark
1
NOTES
0.5
0.5
1
OR
Specific heat: heat per unit mass to raise the
temperature by 1 ℃.
(c)
(i)
1
(ii) m=13kg,
𝑎 = 2 𝑚/𝑠 2 ,
𝑓𝑘 =? ?
∑ 𝐹𝑥 = 𝑚𝑎
0.5
𝑚𝑔𝑐𝑜𝑠53 − 𝑓𝑘 = 𝑚𝑎
OR
𝑚𝑔𝑠𝑖𝑛37 − 𝑓𝑘 = 𝑚𝑎
13 × 9.8 × 𝑐𝑜𝑠53 − 𝑓𝑘 = 13 × 2
𝒇𝒌 = 𝟓𝟎. 𝟔𝟕𝑵
0.5
(ii) 𝜇𝑘 =
𝑓𝑘
𝐹𝑁
𝑭𝑵 = 𝒎𝒈𝒔𝒊𝒏𝟓𝟑 𝑶𝑹 𝑭𝑵 = 𝒎𝒈𝒄𝒐𝒔𝟑𝟕
𝐹𝑁 = 13 × 9.8 × 𝑠𝑖𝑛53 = 101.74𝑁
𝝁𝒌 =
𝟓𝟎. 𝟔𝟕
= 𝟎. 𝟒𝟗𝟖
𝟏𝟎𝟏. 𝟕𝟒
5
(0.5 mark for
FN and fk), and
(0.5 for mg
and its
direction).
0.5
0.5
0.5
0.5
The mark for
equation
Total Marks
Solution
Q# Part
Q3) (a) (i) Work-Energy Principle: The net work done on
an object is equal to the change in the object’s
kinetic energy.
(ii) Electric Current: the rate of flow of charge.
(b)
𝑃 = 4ℎ𝑝 = 4 × 746 = 2984𝑤𝑎𝑡𝑡
𝑡 = 2ℎ = 2 × 3600 = 7200𝑠
𝑊=𝑃𝑡
𝑾 = 𝟐𝟗𝟖𝟒 × 𝟕𝟐𝟎𝟎 = 𝟐. 𝟏𝟒𝟖 × 𝟏𝟎𝟕 𝑱𝒐𝒖𝒍𝒆𝒔
(c)
𝑣1 = 0 , ℎ1 = 35𝑚 , ℎ2 = 20𝑚,
𝐸1 = 𝐸2
1
1
𝑚𝑔ℎ1 + 𝑚𝑣12 = 𝑚𝑔ℎ2 + 𝑚𝑣22
2
2
1 2
9.8 × 35 = 9.8 × 20 + 𝑣2
2
𝒗𝟐 = 𝟏𝟕 𝒎/𝒔
Total Marks
6
7
Mark
1
1
0.5
0.5
1
0.5
𝑣2 =? ?
1
1
0.5
7
NOTES
If student
writes
𝑾𝑵𝒆𝒕 =
∆𝑲𝑬, give
him 0.5
mark.
If student
writes
∆𝑸
𝑰 = ∆𝒕 ,
give him 0.5
mark.
Q# Part
Q4) (a) (i)
(b)
(c)
Solution
(ii)
Mark
0.5, 0.5
𝑣1 = 40 𝑚/𝑠, 𝐴 = 220 𝑚2 ,
𝐹𝑁𝑒𝑡 =? ?
𝐹𝑁𝑒𝑡 = (∆𝑃)𝐴
1
1
𝑃1 + 𝜌𝑔ℎ1 + 𝜌𝑣12 = 𝑃2 + 𝜌𝑔ℎ2 + 𝜌𝑣22
2
2
𝟏
𝟐
𝟐
OR ∆𝑷 = 𝝆𝒂𝒊𝒓 (𝒗𝟏 − 𝒗𝟐 )
𝟐
1
∆𝑃 = × 1.3 × (402 ) = 1040𝑃𝑎
2
𝑭𝑵𝒆𝒕 = (𝟏𝟎𝟒𝟎) × 𝟐𝟐𝟎 = 𝟐𝟐𝟖𝟖𝟎𝟎𝑵
𝑚𝑝𝑖𝑒𝑐𝑒 = 0.250 𝑘𝑔 ,𝑚𝑐𝑎𝑙 = 0.085𝑘𝑔,
𝑚𝑙𝑖𝑞𝑢𝑖𝑑 = 0.300𝑘𝑔
7
0.5 for each
part
1
1
0.5
0.5
0.5
𝐶𝑖𝑟𝑜𝑛 = 450 𝐽/𝑘𝑔. ℃, 𝐶𝐴𝑙 = 4186𝐽/𝑘𝑔. ℃
𝑇1 = 10℃, 𝑇2 = 190℃ ,
𝑇𝑓 = 𝑇𝑒𝑞 = 40℃,
𝐶𝑙𝑖𝑞𝑢𝑖𝑑 = ? ?
𝑄𝑙𝑜𝑠𝑡 = 𝑄𝑔𝑎𝑖𝑛𝑒𝑑
𝑚𝑝𝑖𝑒𝑐𝑒 𝐶𝑖𝑟𝑜𝑛 (𝑇2 − 𝑇𝑓 )
= 𝑚𝑙𝑖𝑞𝑢𝑖𝑑 𝐶𝑙𝑖𝑞. (𝑇𝑓 − 𝑇1 )
+ 𝑚𝑐𝑎𝑙 𝐶𝐴𝑙 (𝑇𝑓 − 𝑇1 )
0.250 × 450 × (190 − 40) =
0.300 × 𝐶𝑙𝑖𝑞. × (40 − 10) + 0.085 ×
900 × (40 − 10)
𝑪𝒍𝒊𝒒𝒖𝒊𝒅 = 𝟏𝟔𝟐𝟎𝑱/𝒌𝒈. ℃
Total Marks
Q# Part
Solution
NOTES
The mark for
conversion
three masses
from gram to kg
1
1
0.5
7
Mark
NOTES
Q5) (a)
𝐼 = 3 𝐴,
𝑡 = 5 𝑚𝑖𝑛
(i) ∆𝑄 =? ?
𝐼=
∆𝑄
∆𝑡
OR
∆𝑄 = 𝐼∆𝑡
∆𝑸 = 𝟑 × 𝟓 × 𝟔𝟎 = 𝟗𝟎𝟎 𝑪
(ii) 𝑁𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 =
𝑁𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 =
(b)
(c)
∆𝑄
𝑒
900
1.6 × 10−19
= 𝟓. 𝟔 × 𝟏𝟎𝟐𝟏
𝐶 = 0.07𝜇𝐹 = 0.07 × 10−6 𝐹
𝑉 = 25𝑘𝑉 = 25 × 103 𝑉
1
PE = C V 2
2
2
1
PE = 0.07 × 10−6 × (25 × 103 )
2
𝐏𝐄 = 𝟐𝟏. 𝟖𝟕𝟓𝐉
𝑄1 = −11𝜇𝐶 = −11 × 10−6 𝐶,
𝑄2 = +4𝜇𝐶 = +4 × 10−6 𝐶,
𝑄3 = −3𝜇𝐶 = −3 × 10−6 𝐶
𝑟23 = 0.2 𝑚,
𝐹3 =? ?
𝑄3 𝑄1
𝐹31 = 𝑘 2
𝑟13
𝐹31
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
The mark for
conversion three
charges from 𝝁𝑪
to 𝑪.
𝑟13 = 0.6 𝑚
1
= (9 × 109 )
𝐹31 = 0.825𝑁
𝐹32 = (9 × 10
0.5
3 × 10−6 × 11 × 10−6
0.62
towards +x direction
9)
0.5
3 × 10−6 × 4 × 10−6
0.22
𝐹32 = 2.7𝑁 towards -x direction
𝐹3 = +𝐹31 − 𝐹32 = 0.825 − 2.7
0.5
𝑭𝟑 = −𝟏. 𝟖𝟕𝟓𝑵
OR 𝑭𝟑 = 𝟏. 𝟖𝟕𝟓𝑵 towards -x direction
Total Marks
0.5
8
7
Direction can be
replaced by +
sign with the
value of the
force.
Direction can be
replaced by - sign
with the value of
the force.