A simple pendulum is used in physics laboratory for experiment to obtain experimental value of gravitational acceleration g. A student measures the length of pendulum 0.51m, displaces it through 10° from equilibrium position and released it. Using a stopwatch, the student measures period of pendulum as 1.44s. Determine the experimental value of the gravitational acceleration. (Ans : g = 9.712 m/s²)
Answers
time period of pendulum , T = 1.44s
Use formula,
Squaring both sides,
T² = 4π²l/g
g = 4π²l/T²
= 4(3.14)² × 0.51/(1.44)²
= (3.142)² × 0.51/(0.36 × 1.44)
≈ 9.712 m/s²
Hence, experimental value of g = 9.712m/s²
Answer:
Poisson's ratio is the ratio of transversal strain. to the corresponding longitudinal strain.
e.g., Poisson's ratio = transversal strain/longitudinal strain
actually, Poisson's ratio is the material constant which is need to quantity elasticity in mathematical terms.
The range of poisson's Ratio is empirically found to be in range of 0–0.5.
for example : Poisson's ratio of gold is 0.42 which lies in interval [0, 0.5].
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
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
abhi178 • ★ Brainly Teacher ★
Physics • 19 hours ago
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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.
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Physics • 19 hours ago