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Q1. The error in two readings of A are 0.01 and -0.03, then mean absolute error is ?
  • 1) 0.02  
  • 2) - 0.02  
  • 3) 0.002  
  • 4) - 0.004  

Solution

The absolute value of the errors is taken. So mean absolute error = (0.01 + 0.03) / 2 = 0.02    
Q2. The mass of a box measured by a grocer's balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is the total mass of the box and account for its accuracy.

Solution

The total mass of the box is equal to the sum of the masses of box and constituents in the box. i.e. M = 2.3 + 0.02015 + 0.02017 = 2.34032 kg Since mass of the box is the least accurate because known to one tenth of kg only, therefore, the result must not be more accurate than one tenth of kg. M = 2.3 kg
Q3. Find the option with 3 significant figures.
  • 1) 0.0268 m  
  • 2) 2.608 cm  
  • 3) 2.068 cm  
  • 4) 26.08 mm  

Solution

0.0268 have three significant figures where as 2.608 cm, 26.08 mm and 2.068 cm have four significant figures.        
Q4. Three physical quantity having dimensions [M L-1 T-2] are
  • 1) Stress, pressure, work
  • 2) Stress, pressure, coefficient of elasticity
  • 3) Stress, pressure, torque
  • 4) Stress, pressure, density

Solution

Pressure = Force/Area = ML1T-2/ L2 = [M L-1 T-2] Stress = Force/Area = ML1T-2/ L2 = [M L-1 T-2] Coefficient of elasticity = Stress/strain = [M L-1 T-2]
Q5. 1 unified atomic mass unit is equal to.
  • 1) 1/13 of the mass of an atom of carbon-13 isotope
  • 2) 1/12 of the mass of an atom of carbon -12 isotope
  • 3) 1/14 of the mass of an atom of carbon-14 isotope
  • 4) 1/13 of the mass of an atom of carbon-12 isotope

Solution

1 unified atomic mass unit is equal to 1/12 of the mass of carbon-12 [12C6] isotope.
Q6. A 5.2 g of a substance occupies 1.11 cm3. Its density to appropriate significant figure is:
  • 1) 4.685  
  • 2) 4.68  
  • 3) 4.7  
  • 4) 4.6845  

Solution

 There are 2 significant figures in measured mass whereas there are only 3 significant figures in measured volume. Hence density should be expressed to only 2 significant figures.        
Q7. Magnitude of force F experienced by a certain object moving with speed v is given by F = Kv3, where K is a constant. The dimensions of K are
  • 1) [ML-3T]
  • 2) [MLT-2]
  • 3) [ML-2T]
  • 4) [M-2LT]

Solution

Q8. 1 fermi is equal to:
  • 1) m
  • 2) m
  • 3) m
  • 4) m

Solution

1fm = 10-15m.
Q9. The velocity  (in cm s-1) of a particle is given in terms of time t (in s) by the equation The dimensions of a, b and c are
  • 1) L T -2, L, T
  • 2) L2, T, L T -2
  • 3) L T 2, L T, L
  • 4) L, L T, T 2

Solution

LT-1 = a T a = L T -2  Similarly,
Q10. The nearest star to our solar system is 4.29 light year away.  How much is this distance in terms of parsecs?

Solution

Q11. The number of significant figures in the distance of one light year , 9.4605 × 1015 m is
  • 1) Four
  • 2) Five
  • 3) Thirty
  • 4) Three

Solution

As the power of 10 is irrelevant in the determination of significant figures hence, the numbers of significant figures are 5.
Q12. The length of the rod as measured in an experiment was found to be 3.23 m, 3.25 m, 3.27 m, 3.22. Find the absolute error.

Solution

 
Q13. derive the expression for viscous force acting on spherical body of radius r moving with velocity v through viscous liquid of co-efficient of viscosity η.

Solution

Q14. Approximate ratio of 1 fermi to 1 angstrom is ..............
  • 1) 10-15m
  • 2) 10-5m
  • 3) 10-9m
  • 4) 10-6m

Solution

1 fermi = 10-15 m, 1angstrom = 10-10 m Hence,1 fermi / 1angstrom  = 10-15/ 10-10 = 10-5 m
Q15. The order of magnitude of a number expressed in scientific notation is
  • 1) all digits in the number
  • 2) number of decimal places
  • 3) number of digit before decimal
  • 4) power of ten

Solution

The order of magnitude of a number expressed in scientific notation is power of ten.
Q16. Which of the quantity is dimensionless?
  • 1) Angle
  • 2) Work
  • 3) Force
  • 4) Area

Solution

Angle = length of Arc/ radius = L/L = 1 = [M0 L0 T0]
Q17. The most suitable instrument for measuring the size of an atom is.
  • 1) Screw gauge
  • 2) Optical Microscope
  • 3) Electron microscope
  • 4) Vernier calliper

Solution

Only electron microscope has resolution of 0.5-0.6 Angstrom which is less than the size of an atom.
Q18. Check the dimensional consistency in the following cases:

Solution

Q19. Angstrom is the unit of
  • 1) Length
  • 2) Light
  • 3) Mass
  • 4) Time

Solution

Ao is the unit of length.
Q20. What is the dimension of (½)at2?
  • 1) length/time2 ([L]/[T2])  
  • 2) length [L]  
  • 3) length/time([L]/[T])  
  • 4) time[T]  

Solution

½ a t2 has the dimension of length since the dimension of acceleration is L/T2 and multiplying it by T2 leaves us with the dimension of length.
Q21. Relative error in Z, given that Z = 1/A2 is:
  • 1) half the error in A  
  • 2) the square of the error in A  
  • 3) two times the error in A  
  • 4) square root of error in A.  

Solution

It is two times the error in A. If two quantities are multiplied or divided the relative error in the result is the sum of the relative errors in the multipliers.    
Q22. Random errors occur due to:
  • 1) parallax  
  • 2) imperfect design of instruments  
  • 3) incorrect rounding off  
  • 4) fluctuations in temperature, supply,error in reading  

Solution

They arise due to random and unpredictable fluctuations in experimental conditions (temperature, supply etc) and personal errors by the observer taking readings.    
Q23. Express one parsec in terms of light year.

Solution

Q24. One angstrom is equal to __________ .
  • 1) 10-12m
  • 2) 10-11m
  • 3) 10-10m
  • 4) 10-9 m

Solution

One angstrom (1o) is equivalent to 10-10m.
Q25. Find the percentage error, if the absolute error is 0.01 and average length of object is 3.25 m.

Solution

   
Q26. Given y = a sin (ωt - k x), where x is position and t is time. Show that ω / k has dimensions of velocity.

Solution

Q27. Find the dimensions of a and b in the Vanderwaal's equation

Solution

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