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Ibeere 1 Ìròyìn
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
Ibeere 2 Ìròyìn
= m(v2−u2)2s
F = 0.05(0−(500)2)2×0.25
F = 25 000N
Ibeere 3 Ìròyìn
Ibeere 4 Ìròyìn
Ibeere 5 Ìròyìn
6 x 105 = V2R t
R = 240×240×5×606×105
= 28.8Ω
Ibeere 6 Ìròyìn
Ibeere 7 Ìròyìn
= Height = 1031.5×102
= 6.67mm
Ibeere 8 Ìròyìn
Awọn alaye Idahun
Ibeere 10 Ìròyìn
Awọn alaye Idahun
Ibeere 11 Ìròyìn
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
Ibeere 13 Ìròyìn
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
Ibeere 14 Ìròyìn
e1 = 20 - L ; e2 = 5cm
f1e1 = f2e2
10020−L = 1005
hence, L = 15cm
Ibeere 15 Ìròyìn
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
Ibeere 16 Ìròyìn
Awọn alaye Idahun
Ibeere 18 Ìròyìn
1.98gs = 1.98 x 1
= 600 cou.
Q = It
600 = 2 x t
t = 300sec = 5 mins
Ibeere 19 Ìròyìn
5002500 = 120Vs
Vs = 600v
Ibeere 20 Ìròyìn
load5 = 1000.5
load = 1000N
Ibeere 21 Ìròyìn
Awọn alaye Idahun
Ibeere 22 Ìròyìn
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
Awọn alaye Idahun
Ibeere 23 Ìròyìn
= 340170
= 2m ; for closed pipe
L = 14 λ
= 14 x 21
= 0.5m
= 50cm
Ibeere 24 Ìròyìn
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
Awọn alaye Idahun
Ibeere 27 Ìròyìn
Awọn alaye Idahun
Ibeere 28 Ìròyìn
Awọn alaye Idahun
Ibeere 31 Ìròyìn
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
Ibeere 32 Ìròyìn
Awọn alaye Idahun
Ibeere 33 Ìròyìn
Awọn alaye Idahun
Ibeere 34 Ìròyìn
V1 = 2V2
1.5×2V2−73+273 = 4.5×V2T2
T2 = 300K = 27oC
Ibeere 35 Ìròyìn
What is the number of neutrons in the Uranium isotope \( {}^{238}_{92}X \)?
=146
Ibeere 36 Ìròyìn
= Nm3 x m31
= Nm = work
Ibeere 38 Ìròyìn
Awọn alaye Idahun
Ibeere 39 Ìròyìn
m x (80 - θ ) x 1.5 = m x (θ - 20) x 1
θ = 56oC
Ibeere 40 Ìròyìn
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
Awọn alaye Idahun
Ibeere 41 Ìròyìn
R = 14.14N
R = F = ma
14.14 = 2 x a
a = 7.07ms-1
Ibeere 43 Ìròyìn
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
Ibeere 44 Ìròyìn
Ibeere 45 Ìròyìn
depth = velocity x time
= 1500 x 0.5
= 750m = 0.75km
Ibeere 47 Ìròyìn
Ibeere 48 Ìròyìn
Ibeere 49 Ìròyìn
R.d = height insideH2Oheight inside liquid
0.8 = 23X
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