UNITS AND MEASUREMENTS Quiz-13

UNITS AND MEASUREMENTS Quiz-13

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1.Given that r=m² sin pt, where t represents time. If the unit of m is N, then the unit of r is

•  N
•  N²
•  N s
•  N² s

Solution
(b) T-ratios are dimensionless. So the unit of r is N².

Q2.The dimensions of farad are

•  M^-1 L^-2 T² Q²
•  M^-1 L^-2 TQ
•  M^-1 L^- T^-2 Q
•  M^-1 L^-2 TQ^-2

Solution
(a) Farad is the unit of capacitance and C=Q/V=[Q/[ML² T^-2 Q^-1 ] ]=M^-1 L^-2 T² Q²

Q3. One light year is defined as the distance travelled by light in one year.
The speed of light is 3×10⁸ms^-1. The same in metre is

•  3×10¹²m
•  9.461×10¹⁵m
•  3×10¹⁵m
•  None of these

Solution

(b) One light year =3×10⁸ m/s year =(3×10⁸)/s×365×24×60×60s
=3×10⁸ ×365×24×60×60m
=9.461×10¹⁵m

Q4. The mean time period of second’s pendulum is 2.00s and mean absolute error in the time period is 0.05s. To express maximum estimate of error, the time period should be written as

•  (2.00±0.01) s
•  (2.00+0.025) s
•  (2.00±0.05) s
•  (2.00±0.10) s

Solution
(c) Mean time period T=2.00 sec & Mean absolute error = ∆T=0.05 sec
To express maximum estimate of error, the time period should be written as (2.00 ±0.05)sec

Q5.A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale
which match with 16 main scale divisions. For this Vernier callipers , the least count is

•  0.02 mm
•  0.05 mm
•  0.1 mm
•  0.2 mm

Solution
 (d) 20 VSD=16 MSD 1 VSD=0.8 MSD

Least count =MSD-VSD =1 mm-0.8 mm=0.2 mm 

Q6.  The relative density of material of a body is found by weighing it first in air and then in water. If the weight
in air is (5.00±0.05) newton and weight in water is (4.00±0.05) newton .Then the relative density along
with the maximum permissible percentage error is

•  5.0±11%
•  5.0±1%
•   5.0±6%
•  1.25±5%

Solution
(a) Weight in air = (5.00±0.05)N
Weight in water =(4.00 ±0.05)N
Loss of weight in water =(1.00±0.1)N
Now relative density=(weight in air)/(weight loss in water) i.e.R.D=(5.00±0.05)/(1.00±0.1)
Now relative density with max permissible error =5.00/1.00±(((0.05/5.00 )+(0.1/1.00 )))×100=5.0±(1+10)%
=5.0±11%

Q7. The dimensions of coefficient of self inductance are

•  [ML² T^-2 A^-2 ]
•  [ML² T^-2 A^-1 ]
•  [MLT^{-2 -2} ]
•  [MLT^{-2 -2} ]

Solution
(a) Energy U=1/2 LI² ⇒ L=2U/I² ∴[L]=[U]/[I]² =[ML² T^-2 ]/[A]²
=[ML²T^-2 A^-2 ]

Q8.If force (F), length (L) and time (T) are assumed to be the fundamental units, then the dimensional
formula of the mass will be

•  [FL^-1 T² ]
•  [FL^-1 T^-2 ]
•  [FL^-1 T^-1 ]
•  [FL²T^-2 ]

Solution
(a) From Newton’s second law Force (F)=Mass (M)×acceleration Dimensions of [F]=[MLT^-2 ]
∴ [M]=[FL^-1 T² ]

Q9. Inductance L can be dimensionally represented as

•  ML² T^-2 A^-2
•  ML² T⁴ A^-3
•  ML T^-2 A^-2
•  ML² T⁴ A³

Solution
(a) E=1/2 Li² hence L=[ML² T^-2 A^-2]

Q10. Linear momentum and angular momentum have the same dimensions in Force =X/(Density ) is given by

•  Mass and length
•  Length and time
•  Mass and time
• Mass, length and time

Solution
(c) Linear momentum =[MLT^-1]
Angular momentum =[ML² T^-1]