305 ii Measure height of I2 and divide by height of object 3 4 5 2004 Place

305 ii measure height of i2 and divide by height of

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305 (ii) Measure height of I2 and divide by height of object 3. 4. 5. 2004: Place candle at a distance as shown below u v
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306 - Adjust until a sharp image is formed - Measure U and V and record - Repeat this procedure for other values of u - For each set of values find using 1 / f = 1 / u + 1 / v - Calculate the average value (b) Draw one ray from top of object straight through the optical centre of the lens a - Draw another ray parallel to the principal axis which should pass through principal axis after refraction - Where they meet, draw the image. (c) Short sightedness 6. (a) (i)
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307 (ii) (b) (i) A-Diaphragm B-Film (ii) The distance between the lens is adjusted so that the image is formed on the film (iii) Shutter- Opens for some given time to alloy rays from the object to fall on the film creating the image impression. A (Diaphragm) controls intensity of light entering the camera B (Film) coated with light sensitive components which react with light to create the impression (c) (i) Magnification = V = 3 U Since v + u = 80 U = 80 v V = 3 80 v V= 240 3v V= 60 cm
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308 (ii) From above u = 20cm I = I + I + I + I F v u 60 20 7. f= 15cm 8. f= 14.3cm 9. Given information 10. u= 16cm f = 12cm Applying l = l + l f u v then I = l + l v f u = 1 1 12 V16 = l = 4 -3 = 1 V 48 48 Hence distance = 48 (ii) Nature: image is real - Inverted - 48 cm from the lens - Magnified (iii) Magnification, m= v = 48 cm = 3 u 16 cm
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309 11. Solution Given information V= -6 cm U = + 15cm Applying I = I + I F u v = I I 15 6 2-5 30 I = -3 = -1 F 30 10 F= -10 F= 10 cm 12. Solution P= I F P = 1 / 10 = 1 / 10/100 m = 1 10 / 100 m = 1 x 100 / 10 Power = 10 dioptre (D)
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310 13. Similarity Differences - Both have converging lens - In the eye, retina act as the screen while in the camera a light sensitive film act as one - Focal length of the eye lens is variable while that of the camera is fixed.
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311 2. UNIFORM CIRCULAR MOTION 1. When tension is max then r= l T= F = Mv 2 r 2 = 0.1 x V2 1.25 V2 = 2 x 1.25 0.1 V= 5m/s 2. (a) (i) & (ii) (b) (i) S= Ut + ½ at 2 t = (2 x 5) 10 = 1 seconds (ii) S = Ut = 30 x 1 = 30m
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312 (iii) V 2 = U 2 + 2as V= √(2 x 10 x 5) = 10m 3. Principal of conical pendulum Or principle of circular motion 4. V= wr But w = θ = 2 x 7 = 14 rads -1 T 1 V= 14 x 0.20 = 8.8 m/s 5. Centripetal acceleration of bucket is equal or higher than gravitational acceleration of the water. 6. (a) Keep angular velocity ω 1 constant; Centripetal force provided by mg; Fix the mass m and measure r; repeat for different values of m. Tabulate the values (b) Force- Mg. Calculate and fill a column for force and another for radius in m Plot the graph
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313 (ii) Gradient of the graph = 0.625 0.1 = 1.167 N 1m 0.525 0.075 Force F, on the body = m b w 2 r Where m b = mass of the body M b w 2 = Gradient of the graph = 1.167 W 2 = 1.167 = 11.67 0.1 W= √11. 67 = 3.42 rads -1 7. 2 f = 2 x 8 = rw 2 = 0.2 (16 ) 2 = 505.3 m/s 8. (a) Angular displacement per second. (b) In 13s, angle turned = 300-170 = 130 rad w= θ = 130 = 10rads-1 t 13 So 300 = 10 t+13 10t + 130 = 300 10t = 170 t = 17 sec
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314 (c) (i) Plot the graph which is a straight line. (ii) From T = (mr)w 2 C and y = mx + c Y = T, grad = mr, y = intercept = -C Obtain gradient and equate it to mr.
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