Q:

In India, the state known for its sandalwood is

A) Karnataka	B) Assam
C) Kerala	D) West Bengal

Answer & Explanation Answer: A) Karnataka

Explanation:

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Filed Under: Indian Economy

35 16168

Q:

When water freezes its density

A) decreases	B) remains constant
C) increases	D) become zero

Answer & Explanation Answer: A) decreases

Explanation:

When water freezes its density decreases.

Water at ordinary temperatures contracts and increases in density as it is cooled, like most substances. But at about 4°C it reaches a maximum density and then decreases in density as it approaches the freezing point.

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Filed Under: Chemistry
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50 16161

Q:

The plateau rich in minerals in india is

A) Malwa plateau	B) Chhota Nagpur Plateau
C) Deccan Plateau	D) Mysuru Plateau

Answer & Explanation Answer: B) Chhota Nagpur Plateau

Explanation:

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Filed Under: Indian Geography

52 16142

Q:

What goes in the water black and comes out red

Answer

A Lobster is one which goes black in water and comes out red.

lobster1585556605.jpg image

Workspace

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45 16139

Q:

When was the first national Forest Policy issued by the Government of India

A) 1940	B) 1952
C) 1942	D) 1999

Answer & Explanation Answer: B) 1952

Explanation:

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Filed Under: Indian Economy

40 16129

Q:

Which sector of Indian Economy contributes largest to the Gross National Product

A) Primary sector	B) Secondary sector
C) Tertiary sector	D) Public Sector

Answer & Explanation Answer: C) Tertiary sector

Explanation:

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16 16109

Q:

Which one of the following elements is essential for construction is nuclear reactors

A) Cobalt	B) Nickel
C) Zirconium	D) Tungsten

Answer & Explanation Answer: C) Zirconium

Explanation:

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59 16098

Q:

The reduction potential of hydrogen half-cell will be negative if :
(1) $P (H_{2})$ = 2 atm and [H+] = 2.0 M
(2) $P (H_{2})$ = 1 atm and [H+] = 2.0 M
(3) $P (H_{2})$ = 1 atm and [H+] = 1.0 M
(4) $P (H_{2})$ = 2 atm and [H+] = 1.0 M

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: D) Option 4

Explanation:

$H^{+} - e^{-} \to \frac{1}{2} H_{2}$

Apply Nernst equation

$E = 0 - \frac{0.059}{1} \log (\frac{P H_{2}^{\frac{1}{2}}}{[H^{+}]})$

$E = - \frac{0.059}{1} \log (\frac{2^{\frac{1}{2}}}{1})$

Therefore E is Negative

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Exam Prep: AIEEE

26 16054

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