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 The correct answer is: (a), (c)

(a), kinetic 

(c) mechanical

 Hold the given piece of glass over some printed number: 

(i) If the letter appear magnified, the given piece is a convex lens. 

(ii) If the letters appear diminished, the given piece is a concave lens. 

(iii) If the letters appear to be of the same size, then it is a plane glass mirror.

The focal length of the lens used as magnifying glass is kept small to increase its magnifying power. 

Focal length depends upon the refractive index of the material of the lens with respect to the medium surrounding it. Since the surrounding medium changes from air to water, the refractive index decreases and so the focal length of the lens increases.

The magnifying power of a simple microscope is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye if it is assumed to be placed at the least distance of distinct vision. Shorter the focal length of the lens used, larger is its magnifying power.

The distance of object (u), the distance of image (u) and focal length if of a lens are related as

1/v - 1/u = 1/f

fl = 25cm = 0.25m

P = 1/0.25 = 4D

 The subjective property of light related to its wavelength is colour.

Ergot fungus, Claviceps purpurea yield LSD.

Correct answer is (d) Claviceps purpurea

Correct answer is (c) hallucinogens

Object is at infinity Image is formed at F₂ on the other side of lens. Image is real; inverted and almost diminished to a point.

Correct answer is (c) Amphetamines – Depressant

(a) It digests protein in acidic medium

Object is between F₁ and O. Image is formed on the same side of lefts. Image is virtual; erect and magnified. 

29°C, 12°C, 0°C, -4°C, – 16°C.

Convex lens.

To form an inverted and enlarged image by a convex lens, the object should be placed between F and 2F infront of the lens.

The image will be real. 

The ray diagram is shown in figure. For the object AB, the images is A’B’.

In a photoelectric experiment, it is observed that when ultraviolet light fell on the emitter metal plate, no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value, called the threshold frequency. This minimum frequency depends on the nature of the material of the emitter plate. 

Average atomic mass of Neon (Ne) =

\(\frac{(At.\,mass\,of\,^{20}Ne \times \%Abundance)+(At.\,mass\,of\,^{21}Ne \times \%Abundance)+(At.\,mass\,of\,^{22}Ne \times \%Abundance)}{100}\)

\(=\frac{(19.99924u \times 90.92)+(20.9940u\times 0.26)+(21.99414u \times 8.82)}{100}\)

= 20.1707 u.

∴ Average atomic mass of neon = 20.1707 u

Correct option (d) r²

Explanation:

The magnetic moment of a current carrying conductor = i.n πr² = niA.

i.e. Magnetic moment ∝ r²

(d) simple harmonic with amplitude √(A²+B²). 

Explanation:

The given equation can be written as, 

x = √(A²+B²){(A/√(A²+B²))sin⍵t+(B/√(A²+B²))cos⍵t}      

Since the magnitudes of A/√(A²+B²) and B/√(A²+B²) are less than 1 and the sum of their square is also 1 we can find an angle α between 0 and 2π such that 

 sinα=B/√(A²+B²) and cosα=A/√(A²+B²) 

So, x = √(A²+B²){sin⍵t.cosα+cos⍵t.sinα} 

→x = √(A²+B²).sin(⍵t+α) 

This is an equation of simple harmonic motion with amplitude √(A²+B²). 

Hence option (d). 

(a) Let the area be A m²

Average power per square meter = 200 W

Useful electrical energy produced per square meter,

= 20/100 x 200 = 40 Watt

Energy required by the family = 8 kW = 8000 watt

Hence, A = 8000/40

= 200 m²

(b) This area is comparable to that of a roof of a house of 14 m x 14 m.

The suitcase is lifted vertically up to the stairs. The porter has a vertical displacement against weight. Hence, he does work.

Power = 40kW

Force = 20,000N

Power = force × velocity

Velocity = Power/Force = 40 KW/20000 = 2 m/s

1. Right hand grip rule.

2. The magnitude, direction of the magnetic fields at A, and C:

• A – outward to the plane of paper
• C-Inward to the plane of paper.

• A concave mirror produces an erect, virtual and magnified image of an object placed between its pole and focus. 
• A dentist uses this principle to get a clear and distinct image of teeth, hence, a dentist uses a concave mirror.

Original length, L = 2πr

Extension, ΔL = 2πR - 2πr = 2π(R - r)

Strain, ΔL/L = 2π(R - r)/2πr = (R - r)/r

Young's modulus, Y = {(F/A)}/{(R - r)/r}

or, F = YA(R - r/r)

Increase in length due to thermal expansion, when temperature of the rod of length L increase by θ, is

ΔL = α L θ;

Strain = ΔL/L = α θ

As Y = Stress/Strain = S/α θ

or Y ∞ 1/α because S and θ are constants for the two rods in this question.

Hence, Y₁/Y₂ = α₂/α₁ = 3/2

The rivers like Ob, Yenisey and Lena which meet the Arctic Ocean remain frozen for most part of the year. So, they are not useful as waterways.

Answer is C. In 1498

The River Nile flows through Uganda, Sudan and Egypt countries. Blue Nile, which is its main tributary flows from Ethiopia and meets it near Khartoum in Sudan. The joint flow of both the rivers flows through Sudan and Egypt to meet the Mediterranean Sea.

Data: L = 3.142 m, N = 1000, I = 5A, 

μ₀ = 4π × 10^-7 T∙m/A

The magnetic induction,

B = μ₀ n| = μ₀ \(\left(\cfrac{N}{L}\right)\)|

= (4π × 10^-7) \((\cfrac{1000}{3.142})(5)=\cfrac{20 \times 3.142 \times 10^{-4}}{3.142}\)

= 2 × 10^-3 T

A thin glass slab is formed when a medium is isolated from its surroundings by two plane surfaces parallel to each other.

Correct option is (C) NI

(b) Distilled water

1. Heat from the sun is absorbed by the water on the surface of ocean, but at deeper levels of ocean, the temperature is very less. 

2. So, there is temperature difference between the water “at the surface of ocean” and at “deep levels”.

3. This difference in temperature is called Ocean Thermal Energy.

(c) hydrogen is evolved at cathode

In Oersted’s experiment, when there is no current in the wire, the magnetic needle is at rest along the north-south direction.

(1) When a current is passed through the wire, the needle is deflected.

(2) When the current through the wire is increased, the deflection of the needle increases.

(3) When the current through the wire is stopped, the needle comes to rest in its original position along the north-south direction.

(4) When the current through the wire is reversed, the needle is deflected in the direction opposite to that in the first case.

(5) When the distance between the magnetic needle and the wire is increased, keeping the current through the wire constant, the deflection of the needle becomes less.

Let the heights of the cones be h and 3h units and the radii of their bases be 3r and r respectively. 

Then, 

their volumes are 

V₁ = \(\frac{1}{3}\)π(3r)²×h and 

V₂ =  π×r²×3h

= \(\frac{V_1}{V_2}\) = \(\frac{3\pi r^2h}{\pi r^2 h}\)

= \(\frac{3}{1}\)

Let the heights of the cones be h and 3h and radii of their bases be 3r and r respectively. Then, their volumes are 

Volume of first cone (V₁) = \(\frac{1}{3}\)π(3r)²h 

Volume of second cone (V₂) = \(\frac{1}{3}\)πr²(3h) 

Now, V₁/V₂ = \(\frac{3}{1}\) 

Ratio of two volumes is 3:1.

Answer is (A) 

We know that a number is divisible by 3 if the sum of its digits is divisible by 3. Now sum of the given six digits is 15 which is divisible by 3. So to form a number of five-digit which is divisible by 3 we can remove either ‘O’ or ‘3’. If digits 1, 2, 3,4, 5 are used then number of required numbers = 5!

If digits 0, 1,2,4, 5 are used then first place from left can be filled in 4 ways and remaining 4 places can be filled in 4! ways. So in this case required numbers are 4 x 4! ways. So, total number of numbers = 120 + 96 = 216

(i) There will be as many ways as there are ways of filling 3 vacant places in succession by the given five digits. In this case, repetition of digits is allowed.

Therefore, the units place can be filled in by any of the given five digits.

Similarly, tens and hundreds digits can be filled in by any of the given five digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125

(ii)

In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits.

Therefore, the number of ways of filling the units place of the three-digit number is 5.

Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

(i) There will be as many ways as there are ways of filling 3 vacant places in succession by the given five digits. In this case, repetition of digits is allowed.

Therefore, the units place can be filled in by any of the given five digits.

Similarly, tens and hundreds digits can be filled in by any of the given five digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125

(ii)

In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits.

Therefore, the number of ways of filling the units place of the three-digit number is 5.

Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

Polyesters are preferred for making dress materials. The main reasons include:

• They have light weight.
• They are easy to wash.
• They do not wrinkle easily.
• They are cheap when compared to dresses made from natural fibres.

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. 

They are as follows: 

1. Regulating water reabsorption through ADH 

2. Electrolyte reabsorption though RAAS 

3. Atrial Natriuretic Peptide

i) Regulating water reabsorption through ADH: 

Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood. 

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells. 

This leads to reduction in urine volume and decrease in osmolarity of blood. 

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.

In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys. 

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus. 

[Note: Hypothalamus is a part of forebrain] 

ii) Electrolyte reabsorption through RAAS: 

Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure. 

iii) Atrial natriuretic peptide (ANP): 

A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na+ and Cl reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na+ in urine) and diuresis.

The answer is (c) abscisic acid

• Abscisic acid forms a layer of abscission. 
• This layer disconnect the living tissue of leaf from the other parts.