3 Ω
2 Ω
1 Ω
6 Ω
Correct answer is A
1 = V/R = E/R+r , if r = 2Ω
V = 3/5 E
=> 3E/5R = E/R+2
i.e 3E(R+2) = 5RE
3RE + 6E = 5RE
i.e 3R + 6 = 5R
6 = 5R - 3R
=> 2r = 6 i.e R = 6/2 = 3Ω
4.5Ω
1.0Ω
8.0Ω
2.0Ω
Correct answer is D
Efficiency = power out put/power input * 100/1
= power out put/ power + that due * 100/1
to internal resistance
=> 75/1 = 12R *100/1
12(R+r)
i.e 75/100 = 6/6+r => 75 (6+r) = 100/*6
i.e 450 + 75r = 600
75r =600 - 450
i.e r = 150/75 = 2Ω
i.e r = 2Ω
2.0 J
4.0 J
12.0 J
2.4 J
Correct answer is A
P = F/A => F = PA = PV/L
Volume V\(_2\) = 6 * 2 x 10\(^{-6}\)
W = PΔV = 2 x 10\(^5\) * (12 x 10\(^{-6}\) - 2 x 10\(^{-6}\))
= 2 x 10\(^5\) 10 x 10\(^{-6}\)
= 2.0J
A string is fastened tightly between two walls 24cm apart. The wavelength of the second overtone is
24cm
16cm
12cm
8cm
Correct answer is B
For a string fixed at both ends and plucked at the middle, the fundamental frequency is given
by f0 = v/λ, where v = velocity of sound, λ = wave length.
At its fundamental, note, the the length of the string is given l = λ/2, => λ0 = 2l, where λ0 = wave length fundamental note. for the first over tone, λ1 = l.
For the second over tone, λ2 = 21/3 = (2 x 24)/3 = 16cm.
i.e. λ2 = 16cm.
2m
4m
1m
3m
Correct answer is C
If F = 85.5Hz, V = 342ms-1,
Then wavelength = V/F = 342/85.5 = 4m
Then if wavelength = 4m
=> wavelength/2 = 4/2 = 2m = NN,
Thus NA = wavelength/4 = 4/4 = 1m.