Calculate the work done in stretching steel wire of length 2m and of cross sectional area 0.0225 mm², when a load of 100 N is applied slowly to its free end. ( Young's modulus of steel = 20 x 10¹⁰ N / m²) (Ans: 2.222 J)
Answers
Length of wire, l = 2m
Load , F = 100N
we know, Y = Fl/A∆l
where F is load , l is length of wire, A is cross sectional area of wire and ∆l is extension of wire.
so, ∆l = Fl/AY
= 100 × 2/(0.0225 × 10^-6 × 20 × 10^10 ) = 4.444 × 10^-2 m
Now, workdone = 1/2 × load × extension of wire
= 1/2 × 100 × 4.444 × 10^-2
= 2.222J
Answer:
Answer:
r = 0.5mm = 0.5 × 10⁻³m
= 5 × 10⁻⁴ m
T = 0.04 N/m
ρ = 0.8 gm/cc
= 0.8x 10 ⁻³/10 ⁻⁶ = 800 kg/m3
θ = 10⁰
g = 9.8 m/s²
T = rh ρ g /2 cos θ
∴ h = 2T cos θ/ r g ρ
∴ h = 2x0.04 cos10/ 5 x10⁻⁴x 800x 9.8
h = 2x 0.04x 0.9848/ 5x 800x 9.8 10⁻⁴
h = 2.01 × 10⁻² m
∴ The height of the capillary rise is 2.01 × 10⁻² m
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Verified answer
Given, length of wire, l = 2m
cross sectional area , A = 10^-4 m²
Load , F = 102kgwt = 102 × 9.8 N
extension of wire, ∆l = 0.1cm = 0.001m
Longitudinal stress = F/A
= 102 × 9.8/(10^-4)
= 999.6 × 10^4 ≈ 1 × 10^7 N/m²
Longitudinal strain. = extension of wire/original length of wire = ∆l/l
= 10^-3/2 = 5 × 10^-4
we know, Young's modulus = longitudinal stress/longitudinal strain
= 1 × 10^7/5 × 10^-4
= 20 × 10^9 N/m²
Physics • 19 hours ago
Answer:
Given :
R = 0.5cm = 0.5 × 10⁻²m
n = No. of drops formed = 10⁶
ρ = 13600 kg/m3 ,
T = 0.465 N/m
W = T∆A
P.E = mgh
Volume of single big drop
V = 4 / 3 π R ³
Volume of single small droplet = 4 /3 π r³
Volume of n droplets = n × 4/ 3 πr³
∴ 4/ 3 πR³ = n 4/ 3 π r³
∴ R³ = nr³
∴ r = R /∛n = 0.5x 10⁻² /∛ 10⁶
∴ r = 0.5 × 10⁻²m /10²
r=0.5 x 10⁻⁴m
The single drop is fallen from height h, hence its P.E. = mgh
But, P.E = Work done due to change in area ...(i)
Change in surface area ∆A = (n4πr²2 – 4πR²)
Also, W = T∆A =
T (n4πr ² – 4πR ² )
W = 4πT (nr² – R² )
∴ eq., (i) becomes,
mgh = 4πT (nr²– R² )
But, m = 4 /3 πR³ρ
∴ 4 /3 πR³ρgh = 4πT (nr² – R² )
R³ gh ρ/3 = T (nr² – R²)
∴ h = 3T (nr² – R² )/ R ³ ρ g
=3x0.465[10 ⁶ (0.5x10⁻⁴)² – (0.5x10⁻²)²]/ (0.5x10⁻²)3x13600x9.8
H=3x0.465[0.25x10⁻²)- (0.25x10⁻⁴)]/0.125x1.36x10⁴ x 9.8
On solving
=3x0.465x25x0.09/12.5x1.36x9.8
H=0.2072m
Physics • 19 hours ago
Explanation:
Given :
T = 435.5 dyne/cm = 0.4355 N/m,
θ = 14
0
ρ = 13600 kg/m³
d = 1 cm
∴ r = 0.5 cm
= 5 × 10⁻³ m
T = rh ρ g/ 2 cos θ
h = 2T cos θ/ r g ρ
∴ h = 2 x0.4355x cos140⁰/ 5 x10⁻³x 13600 x9.8
= 0.8710x cos( 90 + 50 )/ 5x 13.6 x9.8
=0.8710( –sin 50 ) /5x 13.6 x9.8
=–0.8710x 0.7660/ 68.0x 9.8
= – 1.001 × 10⁻³m
∴ h = – 1.001 mm
here Negative sign indicates that mercury level will be lowered by 1.001 mm.
Hence to get correct reading h = 1.001 mm has to added.
∴ h = 1.001 mm
Physics • 19 hours ago
Answer:
Given :
d1 = 1mm
∴ r 1 = d 1/2 = 1/ 2
= 0.5 mm = 0.05 cm
d2 = 1.1 mm
∴ r 2 = d2/ 2 = 1.1/ 2 = 0.55 mm = 0.55 cm
T = 75 dyne/cm
F = Tl
l = 2π (r1 + r2 )
F = Tl
= T × 2π (r1 + r2 )
∴ F = 2πT (r1 + r2 )
= 2 × 3.142 × 75 (0.05 + 0.055)
= 2 × 3.142 × 75 × 0.105
∴ F = 150 × 3.142 × 0.105
∴ F = 49.49 dynes
Physics • 19 hours ago
Answer:
given :
nD = tuning fork D frequency = 340 Hz
nC = tuning fork C frequency=8 – 4 = 4 beats per second.
First nC ± nD = 8 (before filing)
nC ± nD = 4 (after filing)
From given condition nC ± nD = 8
∴ nC ± 340 = 8
∴ nC = 340 + 8 = 348 Hz
or nC = 340 – 8 = 332 Hz
when tuning fork C is filed then nC ± nD = 4
∴ nC ± 340 = 4
∴ nC = 340 + 4 = 344 Hz
or nC = 340 – 4 = 336 Hz
The frequency of tuning fork increases on filing. Hence nC ≠ 344 Hz.
If original frequency of tuning fork C is taken 332 Hz, then on filing both the value 344 Hz, and 336 Hz are greater. Also it produces 4 beats per second with tunning fork D.
∴ frequency of tuning fork C = 332 Hz
∴ nC = 332 Hz.
ANSWER
Physics • 19 hours ago