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. The ratio of the dimension of Planck’s constant and that of moment of inertia is the dimension of
Solution
(a) h/I=[(ML2 T-1/(ML2 )]=[T-1]
(a) h/I=[(ML2 T-1/(ML2 )]=[T-1]
Q2.The equation of a wave is given by
Y = A sinω ((x/υ-k ))
where ω is the angular velocity and υ is the linear velocity. The dimension of k is
where ω is the angular velocity and υ is the linear velocity. The dimension of k is
Solution
(b) From the principle of homogenity (x/v) has dimensions of T
(b) From the principle of homogenity (x/v) has dimensions of T
Q3. The dimensional formula for impulse is
Solution
(a) Impulse=force ×time =[MLT-2 ][T]
=[MLT-1 ]
(a) Impulse=force ×time =[MLT-2 ][T]
=[MLT-1 ]
Q4. The radius of the sphere I (4.3±0.1)cm. The percentage error in its volume is
Solution
(a) Percentage error in radius is 0.1/4.3×100. again, V∝R3
(a) Percentage error in radius is 0.1/4.3×100. again, V∝R3
Q5.Size of universe is about
Solution
(d) Size of universe is about 1026 m=106×(9.46×1015)m =1010ly
(d) Size of universe is about 1026 m=106×(9.46×1015)m =1010ly
Q6. A cube has a side of length 1.2 ×10-2 m. Calculate its volume.
Solution
(a) Volume V=I3=(1.2×10-2 m)3=1.728×106 m3 ∵ length lhas two significance figures. Therefore, the correct answer is
V=1.7×10-6 m3
(a) Volume V=I3=(1.2×10-2 m)3=1.728×106 m3 ∵ length lhas two significance figures. Therefore, the correct answer is
V=1.7×10-6 m3
Q7.The SI unit of momentum is
Solution
(b) mv=kg((m/sec ))
(b) mv=kg((m/sec ))
Q8.A calorie is a unit of heat and equal 4.2 J. Suppose we employ a system of units in which the unit of mass is αkg,
the unit of length is β metre and the unit of time is γ sec. In this new system. 1 calorie =
the unit of length is β metre and the unit of time is γ sec. In this new system. 1 calorie =
Solution
(d) [Calorie]=[ML2 T-2] Comparing with general dimensional formula [Ma Lb Tc ], we get
a=1,b=2,c=-2. n2=4.2[1kg/αkg]1 [1m/βm]2 [1s/γs]-2=4.2α-1 β-2 γ2
(d) [Calorie]=[ML2 T-2] Comparing with general dimensional formula [Ma Lb Tc ], we get
a=1,b=2,c=-2. n2=4.2[1kg/αkg]1 [1m/βm]2 [1s/γs]-2=4.2α-1 β-2 γ2
Q9.Number of base SI unit is
(Given that radius of earth = 6400 km)
Solution
(b)7
(b)7
Q10. Dimensions of bulk modulus are
Solution
(b) Bulk modulus K=(normal stress)/(volumetric strain)
=(F/A)/(-∆V/V) =-FV/(A ∆V) Now, F/A=p ∴ K=pV/∆V As volumetric strain is dimensionless.
∴Dimensions of K=dimensions of normal stress
⇒ [K]=[Ml-1 T-2 ]
(b) Bulk modulus K=(normal stress)/(volumetric strain)
=(F/A)/(-∆V/V) =-FV/(A ∆V) Now, F/A=p ∴ K=pV/∆V As volumetric strain is dimensionless.
∴Dimensions of K=dimensions of normal stress
⇒ [K]=[Ml-1 T-2 ]