The height from which a body is released is numerically equal to the velocity acquired finally. Then that height is
A) g
B) 2g
C) 4g
D) 8g
1 Answers
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A body projected vertically upwards, reaches a height of 180 m. Find out `[g=10ms^(-2)]` (i) intial velocity of body (ii) time taken by body to reach
Correct Answer - (i)`60m//s` (ii) 6 sec (iii) 12 sec (iv )160 m (v)`-20m//s` (vi) 2 sec, 10 sec (vii) 6 sec
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A body is projected at an angle of `45^(@)` with horizontal with velocity of `40sqrt(2)m//s`. Find out (i) Maximum height attained by body (ii) Time o
Correct Answer - (i) 80 m (ii) 8 sec (iii) 320 m (iv) 40 m/s (v) 1 : 1 (vi)`y=x-(x^(2))/(320)m` (vii) 80 m (viii) 60 m
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If the escape velocity of a body on Earth is 11.2 km/s, the escape velocity of the body thrown at an angle 45° with the horizontal will be
Correct option is: (A) 11.2 km/s Escape velocity does not depend on the angle of projection. \(V_{escape} = \sqrt {2gRe} \)
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If the escape velocity of a body on Earth is 11.2 km/s, the escape velocity of the body thrown at an angle 45° with the horizontal will be
2 Answers
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Give reason : The approximate atomic mass in Daltons is numerically equal to the number of nucleons in the atom.
The electrons possess negligible mass. They do not contribute much to the mass of an atom. Therefore, the entire mass of an atom is supposed to be present in the nucleus...
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A body released from height ‘h’ takes time’t’ to reach the ground. After time t/2, its height from the ground is
C) \(\cfrac{3h}4\)
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A body is projected horizontally with a velocity √(29) m/s from a height 10 m, the velocity on reaching the ground ……………… m/s.
Correct option is C) 15
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A body is moving with initial velocity 30 m/s. After certain time it acquires velocity of 40 m/s, then the velocity in middle of its journey is
Correct option is A) 25√2
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On certain amount simple interest is 16/25th part of the amount taken then the time and rate of interest is ……………….. (Numerically same)
Correct option is (A) 8 Let the amount taken be Rs P. \(\therefore\) S.I. = \(\frac{16}{25}\times\) P = \(\frac{16P}{25}\) R = T \(\therefore\) S.I. = \(\frac{PRT}{100}\) \(\Rightarrow\) \(\frac{16P}{25}\) = \(\frac{P\times R\times R}{100}\) \((\because\) R = T) \(\Rightarrow\) \(R^2=\) \(\frac{16P}{25}\times\frac{100}P\) \(=16\times4=64=8^2\) \(\Rightarrow\) R = 8 \(\Rightarrow\) T = R = 8
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If b = 0, c < 0, is it true that the roots of x^2 + bx + c = 0 are numerically equal and opposite in sign? Justify.
Correct answer is Yes
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