concave mirror of focal length 5cm
concave mirror of focal length 10cm
convex mirror of focal length 5cm
convex mirror of focal length 20cm
Correct answer is A
No explanation has been provided for this answer.
3m
5m
6m
10m
Correct answer is C
The image of an object placed in front of a plane mirror is always as far behind the mirror as the object is in front of the mirror. Thus when the object is 4m in front of the mirror, the image will be 4m behind the mirror. But when the mirror is moved in towards the man, (that is the man is now 3m from the mirror), the image distance will again decrease by 1m from behind the mirror. Thus the new image position is 3m behind the mirror. Therefore, the distance between the man and his new image is 6m.
4
3
2
1
Correct answer is B
The lowest note (fundamental note) emitted by any stretched string has a fundamental frequecy fo = V/2l = 40Hz
The 1st overtone = 2fo = 2 x 40 = 80Hz
2nd overtone = 3fo = 3 x 40 = 120Hz
3rd overtone = 4fo = 4 x 40 = 160Hz
4th overtone = 5fo = 5 x 40 = 200Hz
Overtones between 40Hz and 180Hz are 80Hz, 120Hz, 160Hz = 3 Overtones
If a sound wave goes from a cold air region to a hot air region, its wavelength will
increase
decrease
decrease then increase
remain constant
Correct answer is B
If a sound wave goes from a cold air region to a hot air region (which will be less dense), the velocity decreases, and since the frequency is always constant, the wave length decreases with the velocity
0.25 x 10-3rad s-1
0.25 x 10-1rad s-1
5.00 x 102rad s-1
2.50 x 103rad s-1
Correct answer is C
The general wave equation is given by;
y = A Sin(wt - 2πx/Wavelength)
where w = angular velocity
Thus comparing this equation with the equation
y = 0.25 x 10-3 Sin (500t - 0.025x)
That Amplitude = 0.25 x 10-3m
the angular velocity(w) = 500 (the coefficient of t in both equations)
∴ w = 500 rad s-1
= 5.00 x 102rad s-1