A rigid bar of mass M is supported symmetrically by three wires each of length l . Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to
(a) Ycopper /Yiron
(b)
(c)
(d) Yiron/Ycopper .
A rigid bar of mass M is supported symmetrically by three wires each of length l . Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to
(a) Ycopper /Yiron
(b)
(c)
(d) Yiron/Ycopper .
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1 Answer
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This is a multiple choice answer as classified in NCERT Exemplar
Y=
D2=
D=
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If is Poisson’s ratio,
Y = 3K (1 - 2 ) ……… (1)
and Y = 2 ……… (2)
With the help of equations (1) and (2), we can write
dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y
Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.
Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 × (0.5×10? × 10? / 0.1) × (0.04)² = 20×10? ³ v²
0.5 × (5×10²) × 1.6×10? ³ = 20×10? ³ v²
0.4 = 20×10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 20e-3 * v^2 = 10e-3 v^2
4 = 10e-3 v^2
v^2 = 400 => v = 20 m/s
As we know that
If length and diameter both are doubled
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