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Swali 2 Ripoti
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Swali 3 Ripoti
= μ mg
8 = μ x 5 x 10
μ = 0.16
Swali 5 Ripoti
= 340170
= 2m ; for closed pipe
L = 14 λ
= 14 x 21
= 0.5m
= 50cm
Swali 7 Ripoti
5002500 = 120Vs
Vs = 600v
Swali 8 Ripoti
1.98gs = 1.98 x 1
= 600 cou.
Q = It
600 = 2 x t
t = 300sec = 5 mins
Swali 9 Ripoti
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Swali 10 Ripoti
load5 = 1000.5
load = 1000N
Swali 11 Ripoti
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Swali 12 Ripoti
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Swali 13 Ripoti
2T2 = √L12L1
T2 = 2.83sec
Swali 14 Ripoti
m x (80 - θ ) x 1.5 = m x (θ - 20) x 1
θ = 56oC
Swali 15 Ripoti
RQRP = LQAPLPAQ
LQ = 2LP
DQ = 2DP
AP = π (2DP)22
= π DP
AQ = π (DP)22
= 14 π DP
therefore, RQRP = 2LP×πDPLP×14πDP
= 2 : 1
Swali 16 Ripoti
F x d2 = constant
F1 x (d1)2 = F2 x (d2)2
d2 = 2d1
F1 x (d1)2 = F2 x (2d1)2
F2 = 14 F1
Swali 17 Ripoti
R = 14.14N
R = F = ma
14.14 = 2 x a
a = 7.07ms-1
Swali 18 Ripoti
depth = velocity x time
= 1500 x 0.5
= 750m = 0.75km
Swali 20 Ripoti
2 X 10-5 x 2.5 = A2×1.51.5×(−20−20)
A2 = 1.498m2
Swali 22 Ripoti
Two thermo flasks of volume \(V_x\) and \(V_y\) are filled with liquid water at an initial temperature of 0oC. After sometime the temperature were found to be \(\theta_x\), \(\theta_y\), respectively. Given that \(\frac{V_x}{V_y}=2\) and \(\frac{\theta_x}{\theta_y}=\frac{1}{2}\) the ratio of the heat flow into the flask is
Hx = Dx x Vx x Cx x θ x
Hy = Dy x Vy x Cy x θ y
HxHy = Dx×Vx×Cx×θxDx×Vx×Cx×θx
= 2 x 12
= 1
Swali 23 Ripoti
= Nm3 x m31
= Nm = work
Swali 24 Ripoti
R.d = height insideH2Oheight inside liquid
0.8 = 23X
Swali 25 Ripoti
A cell gives a current of 0.15A through a resistance of \(8\ \Omega\) and 0.3A. When the resistance is changed to \(3\ \Omega\) the internal resistance of the cell is
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Swali 26 Ripoti
The graph in the fig above describes the motion of a particle. the acceleration of the particle during the motion is
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Swali 27 Ripoti
In the Fig above, Current I passes through the combination, if the power dissipated in the 5 ohm resistor is 40W, then the power dissipated in the 10 ohm resistor is
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Swali 28 Ripoti
Swali 29 Ripoti
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Swali 30 Ripoti
105 = 10 x h x 1
h = 104
Swali 31 Ripoti
x = (0 x 10) + (12 x 10 x 10 x 10) = 500m
distance in 9s; x - (0 x 9) + (12 x 10 x 9 x 9) = 405m
distance in last is = difference
= 500 - 405
= 95m
Swali 32 Ripoti
What is the number of neutrons in the Uranium isotope \( {}^{238}_{92}X \)?
=146
Swali 33 Ripoti
Two divers G and H are at depths 20m and 40m respectively below the water surface in a lake. The pressure on G is P1 while the pressure on H is P2. If the atmospheric pressure is equivalent to 10m of water, then the value of \( \frac{p_2}{p_1} \) is
p1p1 = 4020
= 2
Swali 34 Ripoti
= m(v2−u2)2s
F = 0.05(0−(500)2)2×0.25
F = 25 000N
Swali 35 Ripoti
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Swali 37 Ripoti
V1 = 2V2
1.5×2V2−73+273 = 4.5×V2T2
T2 = 300K = 27oC
Swali 38 Ripoti
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Swali 40 Ripoti
Swali 41 Ripoti
6 x 105 = V2R t
R = 240×240×5×606×105
= 28.8Ω
Swali 43 Ripoti
= Height = 1031.5×102
= 6.67mm
Swali 44 Ripoti
Two rays of light from a point below the surface of water are equally inclined to each other at \(60^\circ\) in water. What is the angle between the rays when they emerge into air? (Take the refractive index of water to be \(\frac{4}{3}\))
r1 = 41.8o and r2 = 41.8o
angle between them = r1 + r2
= 83.6o
Swali 45 Ripoti
In the Fig above, MN is a light uniform meter rule pivoted at O, the 80cm mark. A load of mass 3.00kg is suspended on the meter rule at L, the 10cm mark. If the rule is kept in equilibrium by a string RP, fixed at P and attached to the rule at R, the 20cm mark, then the Tension T on the string is
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Swali 46 Ripoti
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Swali 47 Ripoti
e1 = 20 - L ; e2 = 5cm
f1e1 = f2e2
10020−L = 1005
hence, L = 15cm
Swali 48 Ripoti
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Swali 49 Ripoti
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Swali 50 Ripoti
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