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.
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Q1. Venus looks brighter than other stars because
Solution
𝐼 ∝ 1/𝑟2
𝐼 ∝ 1/𝑟2
Q2. A ray of light travelling in water is incident on its surface open to air. The angle of incidence isθ, which is less than the critical angle. Then there will be
Q3. If the focal length of objective and eye lens are 1.2 𝑐𝑚 and 3 𝑐𝑚 respectively and the object is put 1.25 𝑐𝑚 away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is
Solution
𝑚∞ = − 𝑣o/𝑢0 × 𝐷/𝑓e From 1/𝑓o = 1/𝑣o − 1/𝑢o ⇒ 1/(+1.2) = 1/𝑣o − 1/(−1.25) ⇒ 𝑣o = 30 𝑐𝑚 ∴ |𝑚∞| = 30/1.25 × 25/3 = 200
𝑚∞ = − 𝑣o/𝑢0 × 𝐷/𝑓e From 1/𝑓o = 1/𝑣o − 1/𝑢o ⇒ 1/(+1.2) = 1/𝑣o − 1/(−1.25) ⇒ 𝑣o = 30 𝑐𝑚 ∴ |𝑚∞| = 30/1.25 × 25/3 = 200
Q4. The refractive index of a piece of transparent quartz is the greatest for
Solution
𝜇 ∝ 1/𝜆 ,𝜆𝑟 > 𝜆𝑣
𝜇 ∝ 1/𝜆 ,𝜆𝑟 > 𝜆𝑣
Q5. Refractive index for a material for infrared light is
Solution
𝜇 ∝ 1/𝜆 .
𝜇 ∝ 1/𝜆 .
Q6. In vacuum the speed of light depends upon
Solution
In vacuum speed of light is constant and is equal to 3 × 10𝑚/𝑠
In vacuum speed of light is constant and is equal to 3 × 10𝑚/𝑠
Q7. A point source of light moves in a straight line parallel to a plane table. Consider a small portion of the table directly below the line of movement of the source. The illuminance at this portion varies with this distance 𝑟
from the source as
Solution
𝐸 = 𝐼 cosθ/𝑟2 ⇒ 𝐸 = 𝐼ℎ/𝑟3
𝐸 = 𝐼 cosθ/𝑟2 ⇒ 𝐸 = 𝐼ℎ/𝑟3
Or 𝐸 ∝
1/𝑟3
Q8. When a plane mirror is placed horizontally on a level ground at a distance of 60𝑚 from the foot of a tower, the top of the tower and its image in the mirror subtend an angle of 90° at the 𝑒𝑦𝑒. The height of the tower will be
Solution
tan45° = ℎ/60 ⇒ ℎ = 60 𝑚
tan45° = ℎ/60 ⇒ ℎ = 60 𝑚
Q9. The distance 𝜈 of the real image formed by a convex lens is measured for various object distance 𝑢 . A graph is plotted between 𝜈 and 𝑢 . Which one of the following graphs is correct?
Solution
Think in terms of rectangular hyperbola
Think in terms of rectangular hyperbola
Q10. A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to from the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object?
Solution
A plano-convex lens behaves as a concave mirror if it’s one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface. Therefore, in this lens two refractions and one reflection occur. Let the focal length of silvered lens is 𝐹. /𝐹 = 1/𝑓 + 1/𝑓 + 1/𝑓m = 2/𝑓 + 1/𝑓m Where 𝑓 = focal length of lens before silvering 𝑓m =focal length of spherical mirror.
A plano-convex lens behaves as a concave mirror if it’s one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface. Therefore, in this lens two refractions and one reflection occur. Let the focal length of silvered lens is 𝐹. /𝐹 = 1/𝑓 + 1/𝑓 + 1/𝑓m = 2/𝑓 + 1/𝑓m Where 𝑓 = focal length of lens before silvering 𝑓m =focal length of spherical mirror.
∴
1/𝐹
=
2/𝑓
+
2/𝑅
…..(𝑖)
(∵ 𝑅 = 2𝑓m
𝑛𝑜𝑤 ,
1 𝑓
= (𝜇 − 1)(1 𝑅₁ - 1 𝑅₂ ) ….(𝑖𝑖)
ℎ𝑒𝑟𝑒, 𝑅₁ = ∞,𝑅₂ = 30 𝑐𝑚
∴
1/𝑓
= (1.5 − 1)(1/∞−1/30)
⟹ 1/𝑓= − 0.5/30= −1/60
⟹ 𝑓 = −60 𝑐𝑚
Hence from eq (i) 1/𝐹 = 2/60 + 2/30 = 6/60
𝐹 = 10 𝑐𝑚
Again given that, Size of object = size of image ⟹ 𝑂 = 𝐼
∴ 𝑚 = −𝑣/𝑢
=
𝐼/𝑂
⟹
𝑣/𝑢
= −1
⟹ 𝑣 = −𝑢
Thus, from lens formula 1/𝐹 = 1/𝑣 − 1/𝑢
1/10 = 1/−𝑢 − 1/𝑢
1/10
= − 2/𝑢
∴ 𝑢 = −20 𝑐𝑚
Hence, to get a real image, object must be placed at a distance 20 cm on the left side of lens