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Answer: (a) tangential

The direction of the electric field at a point of electric field line of force is tangential at that point.

Based on Franklin’s convention, rubber and amber rods are positively charged while the glass rod is positively charged.

Correct answer is (a) assurance

Correct answer is (c) transatlantic

Correct answer is (a) unitary

Correct answer is (a) eatable

During sunrise, the light rays coming from the Sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes. In this journey, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. Since blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour is able to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish early in the morning.

(c) portrayal

(c) dismissal

(d) doubtful

Correct answer is (a) backward

Let the three numbers which are in the ratio 3:7:9 be 3x,7x and 9x.

Now, on subtracting 5 from the second number, we get the

Three numbers as 3x, 7x-5 and 9x

Since, these numbers are in AP, therefore, we have

2(7x-5) = 3x+9x

14x-10 = 12x

2x=10

x=5

Hence, the three original numbers are 3x =15, 7x=35 and 9x =45

2nd Method

Since, these numbers are in AP, therefore, the common differences between them will be same, we have

7x-5-3x = 9x-7x+5

4x-5 = 2x+5

2x = 10

x = 10/2 = 5

x=5

Hence, the three original numbers are 3x =15, 7x=35 and 9x =45

(a) \(\sqrt {\frac {1-cosθ }{2}} = \sqrt {\frac {1-(1-2sin^2\frac θ 2)}{2}} = \sqrt {\frac {2 sin^2\frac θ 2}{2}} = sin \frac θ 2\) - it is True.

(b) \(\frac {1+cosθ }{sinθ } = \frac {1+2cos^2\frac θ 2-1}{2\, sin \frac θ 2 cos \,\frac θ 2}= \frac {2\, cos^2\frac θ 2}{2\,sin \frac θ 2 cos\frac θ 2} = cot\frac θ 2 \neq tan \frac θ 2\) -  It is not true.

(c) \(\sqrt {\frac {1-cos \frac {5\pi}6}{2}} = \sqrt {\frac {1-(1-2\, sin^2 \frac {5\pi}{12})}{2}} = sin \frac {5\pi}{12} \neq cos\frac {5\pi}{12}\) - It is false statement.

(d) \(\sqrt {\frac {1-cos 240°}{2}} = \sqrt {\frac {1-(1-2\, sin^2(\frac {240°}{2})}{2}} = sin\, 120°\) - it is true statement.

(e) cot θ = \(\frac {-12}{5}\)

∴ cos θ = \(\frac {12}{\sqrt{(-12)^2 + 5^2}} = \frac {12}{\sqrt {144 + 25}} = \frac {12}{\sqrt{169}} = \frac {12}{13} (∵ \frac {3\pi}{2}< \theta < 2\pi \) lies in 4th quadrant)

∴ 1 - sin² \(\frac θ2\) = \(\frac {12}{13}\) (∵ cos θ =  1-2 sin² \(\frac θ2\) )

= \(2\, sin^2 \frac θ2 = 1- \frac {12}{13}= \frac {1}{13}\) 

= \(\, sin^2 \frac θ2 = \frac {1}{26}\) 

∴ sin \(\frac θ2\) =  \(\frac {1}{\sqrt{26}} = cosec\frac θ2 = \frac {1}{sin\frac θ2} = \sqrt {26}\) - It is true statement.

(f) cos θ = \(\frac {-15}{17} \)        \(\pi < θ < \frac {3\pi}{4}\)

∴ \(\frac {\pi}{2} < \frac θ2 < \frac {3\pi}{8}\) i.e. \( \frac θ2 \) lies in 2nd quadrant.

Now, 1-2 sin²\( \frac θ2 \) = \(\frac {-15}{17}\) (∵ cosθ = 1 - 2 sin²\( \frac θ2 \) )

=  2 sin²\( \frac θ2 \) = 1 + \(\frac {-15}{17}\) = \(\frac {32}{17}\)

=  sin²\( \frac θ2 \) = \(\frac {16}{17}\) 

∴ sin \( \frac θ2 \) = \(\frac {4}{\sqrt{17}}\) (∵ sinθ is positive in 2nd quadrant)

∴ cos \( \frac θ2 \) =  \(\frac {\sqrt{17-16}}{\sqrt{17}} = \frac {-1}{\sqrt 17}\) (∵ cosθ is negative in 2nd quadrant)

∴ tan \( \frac θ2 \) = \(\frac {sin\fracθ 2 }{cos \frac θ 2} = \frac {\frac {4}{\sqrt{17}}}{\frac {-1}{\sqrt{17}}} = -4\) - it is true statement.

(g) sin θ = \(\frac {5}{13}, 0<\theta < 90°\) 

= 0° < \( \frac θ2 \) < 45°

= All trigonometric ratios are positive for \( \frac θ2 \).

Now, cos θ = \(\sqrt{\frac {13^2-5^2}{13}} = \frac {12}{13}\) 

∴ 1-2 sin² \( \frac θ2 \) = \( \frac {12}{13}\) 

=  2 sin² \( \frac θ2 \) = \(1- \frac {12}{13} = \frac 1{13}\) 

= sin \( \frac θ2 \) = \(\sqrt {\frac {1}{26}} = \frac {1}{\sqrt {26}} = \frac {\sqrt{26}}{\sqrt {26}.\sqrt{26}} = \frac {\sqrt{26}}{26}\) -  It is true statement.

(h) tan θ = \(\frac {-3}{4} \)     \(\frac {\pi}{2}<θ <\pi\)

= \(\frac \pi4 < \frac θ 2 < \frac \pi2\)

= \( \frac θ2 \) lies in first quadrant.

Now, \(\frac {2\, tan \frac θ 2​}{\sqrt {1- tan^2 \frac θ 2}} = \frac {-3}{4}\)  (∵ tan θ = \(\frac {2\, tan θ​}{\sqrt {1- tan^2θ}} \) )

= \(\frac {4\, tan^2 \frac θ 2 ​}{1- tan^2 \frac θ 2} = \frac {9}{16} \) (By squaring both sides)

=  64 tan² \( \frac θ2 \) = 9-9  tan² \( \frac θ2 \) 

= 73  tan² \( \frac θ2 \) = 9

=  tan² \( \frac θ2 \) = \(\frac {9}{73}\)

=  tan \( \frac θ2 \) = \(\frac {3}{\sqrt73} \neq -2\)  - Hence, it is false statement.

(i) sin θ = \(\frac {-1}{3} \)       \(\pi < θ < \frac {3\pi}{2}\)

∴ cos θ = \(\frac {\sqrt{3^2-1^2}}{3} = \frac {-2\sqrt2}{3}\) (∵ cosθ is negative in 3rd quadrant)

Also \(\frac {\pi}{2}< \frac θ 2 = \frac {3\pi}{4}\) 

= \( \frac θ2 \) lies in 2nd quadrant

∴ 1-2 sin² \( \frac θ2 \) = \( \frac {-2\sqrt2}{3}\) 

=  2 sin² \( \frac θ2 \) = \( \frac {1+2\sqrt2}{3}\) = \( \frac {3+2\sqrt2}{3}\) 

= sin \( \frac θ2 \) = \(\sqrt { \frac {3+2\sqrt2}{6}}\) 

& 2 cos²\( \frac θ2 \) -1 = \( \frac {-2\sqrt2}{3}\) 

= 2 cos² \( \frac θ2 \) = 1- \( \frac {-2\sqrt2}{3}\) = \( \frac {3-2\sqrt2}{3}\) 

=  cos \( \frac θ2 \) = - \(\sqrt\frac {3-2\sqrt2}{6}\) 

∴  tan \( \frac θ2 \) = \(\frac {sin\frac θ2}{cos \frac θ2} = \frac {\sqrt{\frac{3+2\sqrt2}{\sqrt6}}}{\frac {-\sqrt {3-2\sqrt2}}{\sqrt6}}\) = \(\frac {\sqrt{3+2\sqrt2}}{\sqrt {3-2\sqrt2}} = \frac {-\sqrt{(3+2\sqrt2)^2}}{\sqrt {9-8}} = 3-2\sqrt2\) - It is true statement

(a) 2 sin \(\frac {2\pi}{9}\) cos \(\frac {\pi}{9}\) 

=  sin (\(\frac {2\pi}{9}\) + \(\frac {\pi}{9}\)) + sin (\(\frac {2\pi}{9}\) - \(\frac {\pi}{9}\))

= sin (\(\frac {3\pi}{9}\)) + sin (\(\frac {\pi}{9}\))

= sin (\(\frac {\pi}{9}\)) + sin (\(\frac {\pi}{9}\))

(b) cos 55° cos 45° 

= \(\frac 12 \) (cos (55° + 45°) + cos (55° - 45°))

= \(\frac 12 \) (cos 100° + cos 10°)

(c) sin 6 θ cos 4 θ 

= \(\frac 12 \) (sin 6 θ + 4 θ) + sin ( 6 θ - 4 θ)

= \(\frac 12 \) (sin 10 θ + 2 θ) 

(d) sin \(\frac {\pi}{9}\) + sin \(\frac {2\pi}{9}\) 

= 2 sin \((\frac {\frac {\pi}{9} + \frac {2\pi}{9}}{2}) cos (\frac {\frac \pi9 -\frac {2\pi}{9}}{2})\) 

= = 2 sin \((\frac {3\pi}{18}) cos ( \frac {-\pi}{18})\)  

(e) sin 55° + sin 45°  

= 2 sin \((\frac {55°+45°}{2}) cos (\frac {55°-45°}{2})\) 

= 2 sin \((\frac {100°}{2}) cos (\frac {10°}{2})\) 

= 2 sin 50° cos 5°

(f) cos \(\frac {3\pi}{2} cos \frac {\pi}{2} \) =  0 x 0 

= 0 (∵ \(cos \frac {\pi}{2} = \) 0 & cos \(\frac {3\pi}{2}\) = 0)

(g) cos \(\frac {3\pi}{2} + cos \frac {\pi}{2} \) 

= 0 + 0 = 0

or

(f) cos \(\frac {3\pi}{2} cos \frac {\pi}{2} \) 

= \(\frac 12 \) ( cos ( \(\frac {3\pi}{2} cos \frac {\pi}{2} \)) + cos ( \(\frac {3\pi}{2} - \frac {\pi}{2} \) ))

= \(\frac 12 \)  ( cos ( \((cos 2\pi + cos \pi) = \) \(\frac 12 \) (1 -1) = 0

(g) cos \(\frac {3\pi}{2} + cos \frac {\pi}{2} \) 

= 2 cos \(\frac {\frac {3\pi}{2}+\frac {\pi}{2}}{2} cos \frac {\frac {3\pi}{2}-\frac {\pi}{2}}{2} \) 

= 2 cos π  cos π/2

= 2 x -1 x 0 = 0

m = 1/2
so -v/u = 1/2

Similarly, 

1/v + 1/u = 1/f

so 1/v - 1/u = 1/-20

v/v - v/u = v/-20 (multiply throughout by v)

1 +1/2 = -v/20

3/2 = -v/20

v = -30

1/2 = -v/u

1/2 = 30/u

u = 60

so the object must be placed 60 cm from the mirror. 

Correct option is C) gave me an earful

Correct option: (a) 3000 kVA

Correct option: (b) Voltage 

Answer is (iv) Step–down transformer decreases the AC voltage

i.e. step down transformer decreases the ac voltage.

D. all of above

The Correct option is  B. charged

D. curved path 

The Correct option is  A. radial

B. electrons

Option : (d) 1.6 x 10⁵ C

A. electrostatic induction

Option : (a) a⁴c

Given,

\(\vec E\) = \(cz^2\hat k\) 

z = a

\(\phi\) = ?

\(\phi\) = \(\int \vec E.d\vec s\)

∵ ds = \(dxdy\,\hat k\)  

\(\phi\) = \(\int(cz^2)\hat k\,dx\,dy\,\hat k\)

\(\phi\) = \(\int\limits_0^zcz^2 dx\,dy\) 

\(\phi\) = \(cz^2\)\(\int\limits_0^zdx\,\int\limits_0^zdy\)

\(\phi\) = \(cz^2\)\([x]_0^z\) \([y]_0^z\)

\(\phi\) = \(cz^2\)[z].[z]

\(\phi\) = cz⁴

\(\phi\) = ca⁴ (∵ z = a given)

∴ Option : (a) is correct

Option : (d) 4.5 x 10^-7 J

Correct answer is (B) A.C. dynamo

In medicine, a loss of blood flow to part of the brain, which damages brain tissue. Cerebrovascular accidents are caused by blood clots and broken blood vessels in the brain. Symptoms include dizziness, numbness, weakness on one side of the body, and problems with talking, writing, or understanding language.

The correct option is (A) Soil.

• The water holding capacity of a soil is a very important agronomical  characteristic. 
• small particles have a much larger surface area than the larger sand particles.
• large surface area of particle allows the soil to hold a greater quantity of water.

a. Mary Quant

The trendy ‘mini length’ was created by designer Mary Quant.

A. No, they cannot form a partnership firm since all of them are minors

D. All of the above

a. Natural resources

There are five main hardware components in a computer system: Input, Processing, Storage, Output and Communication devices.

The temperature of the gas is proportional to the average kinetic energy of its molecules. Faster moving particles will collide with the container walls more frequently and with greater force. This causes the force on the walls of the container to increase and so the pressure increases.