PHY101 Most important MCQ's for final term exam fall 2023 File 1

PHY101
Most Important MCQ's 
Final Term Exam Fall 2023 
                                                                                                                                 Prepared by Amna 

1.  The lowest tone produced by a certain organ comes from a 3.0-m pipe with both ends open. If the speed of sound is 340m/s, the frequency of this tone is approximately:

 ► A. 7Hz 

 ► B. 14 Hz 

 ► C. 28 Hz 

 ► D. 57 Hz page no 53

 Reference:- According to handouts t=wavelength/v And as we know we want to find frequency There we made some changes In this equation 1/t=v/wavelength As 1/t=f so F=v/wavelength

2. To raise the pitch of a certain piano string, the piano tuner:

 ► A. loosens the string 

 ► B. tightens the string 

 ► C. shortens the string page no 54 

 ► D. lengthens the string

Reference:- Hnmm look at this line given in handouts on page 54 “So as the observer runs towards the source, she hears a higher frequency ( Higher pitch)” Means to say jitna hum source k kareeb jate jain gin distance kam hota jay ga Orr utni hi ziada frequency ya pitch sunai de gi isi tra jitna piano k string ko kam Karin gin utni ziada pitch produce hogi.

3. Question No: 3 ( Marks: 1 ) -Please choose one A force of 5000N is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 cm and a Young's modulus of 125 × 108 N/m2 . The elongation of the rod is: 

 ► 0.0020mm 

 ► 0.0040mm

 ► 0.14mm 

 ► 0.55mm

Reference:- Page no :48 The elongation of an object is: d = FL/AY Where F is the force, L is the length, A is the cross-sectional area, and L is the modulus of elasticity (Young's modulus). D or del L is obviously the elongation. Be sure your units are correct for doing this, because they need to cancel out to get a final unit of just meters. You want newtons, meters, square meters, and newtons/m^2.

d = (5000 * 5)/(3.14*0.34^2 * 125*10^8) d = 5.5*10^-6 m d = 0.0055 mm So it will elongate 0.0055 mm. I Since D is closest, but is off by 100 times, I suspect that for the Young's modulus, you meant 1.25*10^8, not 125*10^8. I would choose option D.

4. A particle oscillating in simple harmonic motion is:

► never in equilibrium because it is in motion 

 ► never in equilibrium because there is always a force

 ► in equilibrium at the ends of its path because its velocity is zero there

 ► in equilibrium at the center of its path because the acceleration is zero there

5. In simple harmonic motion, the restoring force must be proportional to the: 

 ► amplitude 

 ► frequency 

 ► velocity 

 ► displacement page no:44 

Reference:- Now we have a restoring force that is proportional to the distance away from the equilibrium point

6. A 160-N child sits on a light swing and is pulled back and held with a horizontal force of 100 N. The magnitude of the tension force of each of the two supporting ropes is: 

 ► 60N 

 ► 94N 

 ► 120N

► 190N 

 Question No: 7 ( Marks: 1 ) - Please choose one An object attached to one end of a spring makes 20 vibrations in 10 s. Its angular frequency is: 

 ► 12.6 rad/s 

 ► 1.57 rad/s 

 ► 2.0 rad/s 

 ► 6.3 rad/s 

Reference:- Page no 42 V=20 , t=10 Ω=2.pie.v =2*3.14*20=125.6(1/t)=12.6rad/se

Question No: 8 ( Marks: 1 ) - Please choose one For an object in equilibrium the net torque acting on it vanishes only if each torque is calculated about: 

 ► the center of mass 

 ► the center of gravity

 ► the geometrical center 

 ► the same point 

Reference:- http://myfizika.ucoz.com/_ld/0/48_Test_Bank_12.pdf 

Question No: 9 ( Marks: 1 ) - Please choose one Ten seconds after an electric fan is turned on, the fan rotates at 300 rev/min. Its average angular acceleration is: 

 ► 3.14 rad/s2 

 ► 30 rad/s2 

 ► 30 rev/s2 

 ► 50 rev/min2

Reference:- Page no 28 Avg angular acceleration=(final angular velocity-intial angular velocity)/time =(300-0)/10=30rev/s2

Question No: 10 ( Marks: 1 ) - Please choose one A 4.0-N puck is traveling at 3.0m/s. It strikes a 8.0-N puck, which is stationary. The two pucks stick together. Their common final speed is: 

 ► 1.0m/s 

 ► 1.5m/s 

 ► 2.0m/s 

 ► 2.3m/s 

 Reference:- The case you're describing is called an inelastic collision. Two objects collide, stick to each other and continue their motion as one body. Due to momentum conservation principle, sum of two bodies momenta before collision has to be equal to momentum of the one body after collision. pbefore = pfirst + psecond = m1v1 + m2v2 pafter = (m1 + m2)vcommon Since pbefore = pafter, (m1 + m2)vcommon = m1v1 + m2v2 We can get vcommon from that: vcommon = (m1v1 + m2v2) / (m1 + m2)

11. An object moving in a circle at constant speed 

 ► must have only one force acting on it 

 ► is not accelerating 

 ► is held to its path by centrifugal force 

 ► has an acceleration of constant magnitude

 Reference:- Page no: 29 Now consider a particle going around a circle at constant speed. You might think that constant speed means no acceleration. Bu this is wrong! It is changing its direction and accelerating. This is called "centripetal acceleration", meaning acceleration directed towards the centre of the circle

Question No: 12 ( Marks: 1 ) - Please choose one A plane traveling north at 200m/s turns and then travels south at 200m/s. The change in its velocity is: 

 ► 400m/s north 

 ► 400m/s south 

 ► zero 

 ► 200m/s south 

Reference:- 400 m/s south because it need 200 to overcome the 200 north then another 200 to get going 200 south.

 Question No: 13 ( Marks: 1 ) - Please choose one At time t = 0 a car has a velocity of 16 m/s. It slows down with an acceleration given by −0.50t, in m/s2 for t in seconds. It stops at t = 

 ► 64 s 

 ► 32 s 

 ► 16 s 

 ► 8.0 s

 Reference:- Vf=vi+at Vf=16-0.50t(0) Vf=16 As we know a=final velocity/t And t=final velocity/a 16/.5 = t 32 = t

Question No: 14 ( Marks: 1 ) - Please choose one 1 mi is equivalent to 1609 m so 55 mph is: 

 ► 15 m/s 

 ► 25 m/s 

 ► 66 m/s 

 ► 88 m/s 

Reference:- 1 mile = 1609 meters or 1.609 kilometers so, 55 miles = 1609x55 = 88495 meters or 88.495 kilometers