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Ajụjụ 2 Ripọtì
P2 = 2P ; T1 = 2T2
therefore, P1×10002T2 = 2P2×V2T2
= 25cm
Ajụjụ 3 Ripọtì
mass = velocity x density
= 20 x 19
= 380g
Ajụjụ 5 Ripọtì
PQ is a thin rod on a horizontal table, RS is a plane mirror inclined at \(45^\circ C\) to the horizontal as shown above. The image of PQ as seen in the mirror by the eye T is
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
mcΩ = I2RTt
1000 x c x 10 = 3 x 3 x 20 x 10
0.18Jg-1oC
Ajụjụ 8 Ripọtì
In a resonance tube experiment, a tube fixed length is closed at one end and several tuning forks of increasing frequency used to obtain resonance at the open end. If the tuning fork with the lowest frequency which gave resonance had a frequency \(f_1\) and the next tuning fork to give resonance had a frequency \(f_2\), find the ratio \(\frac{f^2}{f^2}\).
Fλ = velocity
F1 x λ1
= F2 x
λ2
; F1 x 4λ1
= F2 x -λ2
= F2F1 = 3
Ajụjụ 9 Ripọtì
Ajụjụ 11 Ripọtì
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
In the Figure above, QR is a vertical conductor and the current I flows from R to Q, P is a point on the horizontal plane and is to the south of the wire. The direction of the magnetic field at P due to the current is
Ajụjụ 13 Ripọtì
= 50 x 10 x 5 = 500W
Ajụjụ 14 Ripọtì
1.5 = 3×104speed in glass
speed in glass = 2 x 108ms-1
Ajụjụ 15 Ripọtì
= m(u−v)t
16 = 4(v−0)5
v = 20ms-1
Ajụjụ 17 Ripọtì
echo time = 1 ÷ 2 = 0.5 sec.
= 0.5 x 250
= 125m
Ajụjụ 18 Ripọtì
3 x 2 x 10-5 = V2−1010×100
V2 = 10.06cm3
Ajụjụ 19 Ripọtì
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
In a resonance tube experiment, a tube of fixedlength in closed at one end and several turningforks of increasing frequency used to obtainresonance at the open end. If the turning fork withthe lowest frequency which gave resonance had a frequency f1 and the next turning fork to giveresonance had a frequency f2, find the ratio f2/ f1.
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
= 0.99A
IR = 10 x 10 x 10-3 x 5
R = 599 Ω
Ajụjụ 23 Ripọtì
The image above shows an inverted U-tube with the open end, O of one limb below the level, W of water in the tank, in order that water should begin to flow from the tank it is is necessary that
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
= 5 x 3600 x 0.126 x 10-6
= 2.268 x 10-3kg
Ajụjụ 25 Ripọtì
= 5 x 10-4kg
Density = massarea x thickness
500 = 5×10−70.5 x thickness
thickness = 2 x 10-9m
Ajụjụ 26 Ripọtì
The speed of light in vacuum is \(3.0 \times 10^{8}\,\mathrm{ms}^{-1}\). If the refractive index of a transparent liquid is \(\frac{4}{3}\), then the speed of light in the liquid is
43 = 3×108v
v = 2.25 x 10 8ms-1
Ajụjụ 27 Ripọtì
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
Ajụjụ 30 Ripọtì
Which of the following may be used to determine the relative humidity in a physics laboratory I. Manometer
II. Wet-and-Dry Bulb Hygrometer
III. Hair Hygrometer
IV. A Hydrometer
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
Ajụjụ 34 Ripọtì
Each of the figures in the diagram above represent two current carrying conductors situated close to each other. In which two diagrams, are the forces between the two wires attractive
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
e1 = 14 - L and e2 = 16 - L
Using, m1e1 = m2e2
2014−L = 3016−L
L = 10cm
Ajụjụ 36 Ripọtì
Ajụjụ 37 Ripọtì
Ajụjụ 38 Ripọtì
The diagram above shows three circuits, the internal resistances of the batteries are negligible. which of the currents is the largest
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
Density x pressure = constant
D1 x P1 = D2 x P2
13.6 x 0.76 = 1 x P2
P2 = 10m
Ajụjụ 40 Ripọtì
Ajụjụ 41 Ripọtì
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
Akọwa Nkọwa
Ajụjụ 43 Ripọtì
Which of the following figures gives the correct resultant R of two vectors P and Q?
Akọwa Nkọwa
Ajụjụ 44 Ripọtì
I = 2R ; I = 0.2r
24 = 0.2r
r = 0.4Ω
Ajụjụ 45 Ripọtì
Akọwa Nkọwa
Ajụjụ 46 Ripọtì
Ajụjụ 47 Ripọtì
Which of the following may be used to determine relative humidity in a physics laboratory?
I. Manometer
II. Wet-and-dry bulb hygrometer
III. Hair hygrometer
IV. A hydrometer.
Ị ga-achọ ịga n'ihu na omume a?