UPSC CISF-AC-EXE Archives

41. The image formed by a plane mirror is :

The image formed by a plane mirror is :

always virtual and erect.

always virtual but erect or inverted upon the size of the object.

never virtual but always erect.

always real and erect.

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A plane mirror is a flat reflective surface. The image formed by a plane mirror has specific characteristics:
1. **Virtual:** The image is formed where the light rays appear to diverge from, but do not actually intersect. It cannot be projected onto a screen.
2. **Erect:** The image is upright, meaning it is oriented the same way as the object (top is top, bottom is bottom).
3. **Laterally Inverted:** The image is flipped left-to-right.
4. **Same size:** The size of the image is equal to the size of the object.
5. **Same distance:** The image is located behind the mirror at the same distance as the object is in front of the mirror.

Based on these properties, the image formed by a plane mirror is always virtual and always erect.
Option A states “always virtual and erect,” which matches the known properties.
Option B is incorrect because the image is always erect, not dependent on object size.
Option C is incorrect because the image is always virtual.
Option D is incorrect because the image is always virtual, not real.

– A plane mirror produces a virtual image.
– A plane mirror produces an erect image.
– The image is also laterally inverted and the same size as the object.

Real images are formed when light rays actually converge at a point and can be projected onto a screen (e.g., image formed by a projector lens). Virtual images cannot be projected (e.g., image in a plane mirror or through a magnifying glass). Erect means the image orientation is the same as the object; inverted means it is upside down.

42. Dioptre is the SI unit of :

Dioptre is the SI unit of :

size of the lens.

curvature of the lens.

power of the lens.

magnification produced by a lens.

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Dioptre (symbol D) is the unit of measurement for the optical power of a lens or curved mirror.
Optical power (P) of a lens is defined as the reciprocal of its focal length (f) when the focal length is measured in metres.
P = 1 / f (where f is in metres)
If the focal length f is in metres, the unit of power P is m⁻¹, which is defined as the dioptre (D).
For example, a lens with a focal length of +0.5 metres has a power of P = 1 / 0.5 m = +2 D.
A lens with a focal length of -0.25 metres has a power of P = 1 / -0.25 m = -4 D.

– Dioptre is the unit used to measure the optical power of lenses.
– Optical power is the reciprocal of the focal length measured in metres.

Size of the lens is measured in units of length (e.g., diameter in cm). Curvature of the lens is related to the radii of curvature of its surfaces, usually measured in reciprocal metres (m⁻¹), and contributes to the power of the lens along with the refractive index of the material. Magnification is a ratio of image size to object size, which is a dimensionless quantity or expressed as ‘x’. Dioptre specifically quantifies the power of the lens to converge or diverge light.

43. Which one of the following materials cannot be used to make a lens ?

Which one of the following materials cannot be used to make a lens ?

Water

Glass

Plastic

Clay

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A lens is an optical device that transmits and refracts light, causing the convergence or divergence of light rays. For a material to be used to make a lens, it must be transparent or at least translucent enough to allow light to pass through it and must have a refractive index different from the surrounding medium (usually air) so that refraction occurs.
Let’s examine the given options:
A) Water: Water is transparent and has a refractive index. Lenses can be made from water (e.g., water-filled lenses) or exist naturally (e.g., a droplet of water can act as a lens).
B) Glass: Glass is a common, transparent material with a suitable refractive index used extensively for making lenses.
C) Plastic: Many types of plastic are transparent and are widely used for making lenses, especially in eyeglasses, contact lenses, and camera lenses.
D) Clay: Clay is typically opaque (does not allow light to pass through) and therefore cannot be used to make a functional lens that refracts light in the required manner. While it can be molded, its opaqueness makes it unsuitable for transmitting images.

– A lens works by refracting light, which requires the material to be transparent or translucent.
– The material must have a different refractive index from the surrounding medium.
– Opaque materials cannot be used to make standard lenses.

Materials like glass and plastic are commonly used because they are transparent, can be precisely shaped, and have stable optical properties. Water’s shape changes easily, making it less practical for stable lenses unless contained. Clay’s opaqueness fundamentally prevents it from being used for light transmission and refraction in the way a lens operates.

44. The blue colour of the sky and the reddening of the Sun at sunrise and

The blue colour of the sky and the reddening of the Sun at sunrise and sunset are caused due to the phenomenon of :

dispersion of light.

reflection and refraction of light.

aberration of light.

scattering of light.

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The blue colour of the sky and the reddening of the Sun at sunrise and sunset are classic examples of the phenomenon of scattering of light by particles in the atmosphere.
When sunlight enters the Earth’s atmosphere, it interacts with the gas molecules (mainly Nitrogen and Oxygen) and tiny particles present. This interaction causes the light to be scattered in various directions. This process is called scattering.
Rayleigh scattering explains why the sky is blue. It states that the intensity of scattered light is inversely proportional to the fourth power of the wavelength. Shorter wavelengths, like blue and violet light, are scattered much more effectively than longer wavelengths, like red and orange light. When we look at the sky during the day, we see the scattered blue light from all directions.
At sunrise and sunset, the sunlight travels a much longer path through the atmosphere to reach our eyes. During this long journey, most of the shorter wavelength blue and green light is scattered away by the atmospheric particles. The longer wavelength light, such as red and orange, which is scattered less, is left to reach our eyes, making the Sun and the sky around it appear reddish or orange.

– The colour of the sky and the Sun’s colour at sunrise/sunset are due to the interaction of sunlight with atmospheric particles.
– This interaction involves scattering of light.
– Shorter wavelengths (blue) are scattered more effectively than longer wavelengths (red) by atmospheric gases (Rayleigh scattering).
– The path length of sunlight through the atmosphere affects which colours are predominantly seen.

Dispersion of light is the splitting of white light into its constituent colours (spectrum) due to the dependence of refractive index on wavelength (e.g., prism). Reflection is the bouncing of light off a surface. Refraction is the bending of light as it passes from one medium to another. Aberration refers to defects in image formation by optical systems. None of these fully explain the blue sky or red sunrise/sunset phenomenon, which is primarily a scattering effect.

45. Suppose, ‘A’ can complete a job in 10 days, and ‘A’ and ‘B’ together c

Suppose, ‘A’ can complete a job in 10 days, and ‘A’ and ‘B’ together can complete the same job in 6 days. In how many days can ‘B’ alone complete the job ?

15 days

12 days

18 days

20 days

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Let the total amount of work be W.
A completes the job in 10 days. A’s work rate per day = W / 10.
A and B together complete the job in 6 days. Their combined work rate per day = W / 6.
Let B alone take x days to complete the job. B’s work rate per day = W / x.

The combined work rate of A and B is the sum of their individual work rates:
(A’s rate) + (B’s rate) = (A+B)’s rate
(W / 10) + (W / x) = (W / 6)

Since W represents the same job and is non-zero, we can divide the entire equation by W:
1 / 10 + 1 / x = 1 / 6

Now, solve for 1/x:
1 / x = 1 / 6 – 1 / 10

To subtract the fractions, find a common denominator for 6 and 10. The least common multiple is 30.
1 / 6 = 5 / 30
1 / 10 = 3 / 30

So, 1 / x = 5 / 30 – 3 / 30
1 / x = (5 – 3) / 30
1 / x = 2 / 30
1 / x = 1 / 15

Therefore, x = 15.
B alone can complete the job in 15 days.

– Represent work rates as the reciprocal of the time taken (assuming the total work is 1 unit or W).
– The combined work rate is the sum of individual work rates.
– Set up an equation based on work rates and solve for the unknown time.

This type of problem can also be approached by considering “units of work”. If the LCM of 10 and 6 is 30, assume the total work is 30 units.
A does 30 units in 10 days, so A’s rate = 30/10 = 3 units/day.
A and B together do 30 units in 6 days, so their combined rate = 30/6 = 5 units/day.
B’s rate = (A+B)’s rate – A’s rate = 5 units/day – 3 units/day = 2 units/day.
Time taken by B alone = Total Work / B’s rate = 30 units / (2 units/day) = 15 days.
Both methods yield the same result.

46. ‘B’ is 28 years older than ‘A’. After 12 years from now, the age of ‘B

‘B’ is 28 years older than ‘A’. After 12 years from now, the age of ‘B’ will be twice the age of ‘A’. The present age of ‘A’ is :

10 years

12 years

30 years

16 years

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Let the present age of A be A years.
Let the present age of B be B years.
According to the first statement, ‘B’ is 28 years older than ‘A’:
B = A + 28 (Equation 1)

After 12 years from now:
Age of A will be A + 12 years.
Age of B will be B + 12 years.

According to the second statement, the age of ‘B’ will be twice the age of ‘A’ after 12 years:
B + 12 = 2 * (A + 12) (Equation 2)

Now, we substitute Equation 1 into Equation 2:
(A + 28) + 12 = 2 * (A + 12)
A + 40 = 2A + 24

Now, we solve for A:
40 – 24 = 2A – A
16 = A

So, the present age of A is 16 years.
Let’s verify:
If A = 16, then B = 16 + 28 = 44.
After 12 years, A’s age = 16 + 12 = 28.
After 12 years, B’s age = 44 + 12 = 56.
Is B’s age (56) twice A’s age (28)? Yes, 56 = 2 * 28. The condition is satisfied.

– Represent the unknown ages with variables.
– Translate the given information into algebraic equations.
– Solve the system of equations to find the unknown age.

This is a typical age-based problem that can be solved using simultaneous linear equations. Setting up the equations correctly based on the time references (present, after 12 years) is crucial.

47. A solid metallic sphere of 2 cm radius is melted and converted to a cu

A solid metallic sphere of 2 cm radius is melted and converted to a cube. The side of the cube is approximately equal to :

2.4 cm

2.8 cm

3.2 cm

3.6 cm

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When a solid metallic sphere is melted and converted into a cube, the volume of the material remains constant.
Radius of the sphere (r) = 2 cm.
Volume of the sphere (V_sphere) = (4/3)πr³.
V_sphere = (4/3) * π * (2 cm)³ = (4/3) * π * 8 cm³ = (32/3)π cm³.
Let the side of the cube be ‘s’ cm.
Volume of the cube (V_cube) = s³.
Since the volume is conserved, V_cube = V_sphere.
s³ = (32/3)π.
To find ‘s’, we take the cube root: s = ³√[(32/3)π].
Using the approximate value of π ≈ 3.14159:
(32/3)π ≈ (10.6667) * 3.14159 ≈ 33.51 cm³.
s ≈ ³√33.51 cm.
Let’s evaluate the cube of the given options:
A) (2.4)³ ≈ 13.824
B) (2.8)³ ≈ 21.952
C) (3.2)³ ≈ 32.768
D) (3.6)³ ≈ 46.656
The value 3.2³ = 32.768 is closest to 33.51.

– The volume of the material is conserved during melting and recasting.
– Formula for the volume of a sphere: V = (4/3)πr³.
– Formula for the volume of a cube: V = s³.
– Solve for the side ‘s’ by equating the volumes and taking the cube root.

The value of π is an irrational number, so the exact value of ‘s’ is ³√[(32/3)π]. The question asks for an approximate value, so we use the numerical value of π. The calculation shows that 3.2 cm is the closest approximation among the given options.

48. The sum of the numbers 3, 6, 9, 12, … up to the 20 th term is :

The sum of the numbers 3, 6, 9, 12, … up to the 20^th term is :

600

1260

630

960

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The given sequence is 3, 6, 9, 12, … which is an arithmetic progression (AP).
The first term (a) is 3.
The common difference (d) is 6 – 3 = 3 (or 9 – 6 = 3, etc.).
We need to find the sum of the first 20 terms (n = 20).
The formula for the sum of the first n terms of an AP is S_n = n/2 * [2a + (n-1)d].
Substituting the values:
S₂₀ = 20/2 * [2 * 3 + (20 – 1) * 3]
S₂₀ = 10 * [6 + 19 * 3]
S₂₀ = 10 * [6 + 57]
S₂₀ = 10 * 63
S₂₀ = 630

– Recognize the sequence as an Arithmetic Progression (AP).
– Identify the first term (a) and the common difference (d).
– Use the formula for the sum of the first n terms of an AP: S_n = n/2 * [2a + (n-1)d].

Alternatively, the sequence is 3 * 1, 3 * 2, 3 * 3, …, 3 * 20.
The sum is 3 * (1 + 2 + 3 + … + 20).
The sum of the first n natural numbers is n(n+1)/2.
So, the sum of 1 to 20 is 20(20+1)/2 = 20 * 21 / 2 = 10 * 21 = 210.
The sum of the sequence is 3 * 210 = 630.
This confirms the result obtained using the AP sum formula.

49. In a group of 30 students, each student has opted for at least one of

In a group of 30 students, each student has opted for at least one of the two subjects, Hindi and English. Twelve of them have opted for Hindi and twenty-two have opted for English. The number of students who have opted for only English is :

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Let H be the set of students who opted for Hindi and E be the set of students who opted for English. We are given the total number of students, $|H \cup E| = 30$ (since each student opted for at least one subject). We are given $|H| = 12$ and $|E| = 22$. The number of students who opted for both subjects is the intersection of H and E, denoted by $|H \cap E|$. Using the principle of inclusion-exclusion for two sets, we have $|H \cup E| = |H| + |E| – |H \cap E|$. Substituting the given values, $30 = 12 + 22 – |H \cap E|$. This gives $30 = 34 – |H \cap E|$, so $|H \cap E| = 34 – 30 = 4$. The number of students who opted for only English is the number of students in E minus the number of students in both H and E. So, number of students who opted for only English = $|E| – |H \cap E| = 22 – 4 = 18$.

– Total students = Students who opted for Hindi only + Students who opted for English only + Students who opted for both.
– $|H \cup E| = |H \text{ only}| + |E \text{ only}| + |H \cap E|$.
– $|H| = |H \text{ only}| + |H \cap E|$.
– $|E| = |E \text{ only}| + |H \cap E|$.

From $|H \cap E|=4$, we can also find the number of students who opted for only Hindi: $|H \text{ only}| = |H| – |H \cap E| = 12 – 4 = 8$.
Check: Total students = (Only Hindi) + (Only English) + (Both) = 8 + 18 + 4 = 30, which matches the given information.

50. If the sum of five consecutive even numbers is equal to the product of

If the sum of five consecutive even numbers is equal to the product of first five natural numbers, then which one of the following is the largest of those even numbers ?

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The product of the first five natural numbers (1, 2, 3, 4, 5) is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Let the five consecutive even numbers be represented as $x-4, x-2, x, x+2, x+4$, where $x$ is the middle number. The sum of these five consecutive even numbers is $(x-4) + (x-2) + x + (x+2) + (x+4) = 5x$. We are given that this sum is equal to the product of the first five natural numbers, so $5x = 120$. Solving for $x$, we get $x = 120 / 5 = 24$. The five consecutive even numbers are $24-4=20$, $24-2=22$, $24$, $24+2=26$, and $24+4=28$. The largest of these even numbers is 28.

– First five natural numbers are 1, 2, 3, 4, 5. Their product is 120.
– Consecutive even numbers differ by 2. Representing them around the middle term simplifies the sum calculation.
– If the middle term is $x$, the sum of five consecutive even numbers is $5x$.

Using $x-4, x-2, x, x+2, x+4$ simplifies the sum calculation as the constant terms cancel out. Alternatively, if the first even number is $a$, the numbers are $a, a+2, a+4, a+6, a+8$. Their sum is $5a+20$. So $5a+20 = 120$, $5a = 100$, $a=20$. The numbers are 20, 22, 24, 26, 28. The largest is $a+8 = 20+8 = 28$.