A Tortoise A argues with argue with tortoise B according to A we weigh lens in water than Air but B oppose it. Who is correct A or B and which principal use by correct one?
Q. A Tortoise A argues with argue with tortoise B according to A we weigh lens in water than Air but B oppose it. Who is correct A or B and which principal use by correct one?
Ans: Buoyant force is directly proportional to the density of fluid in which a solid is immersed. Water is much denser than air. So more buoyant force act in water Hence , Statement of A is correct.
Q. What is reverberation? Write its one advantage and one disadvantage. How reverberation is reduced?
Ans: The repeated multiple reflections of sound in any big enclosed space, is called reverberation.
When a sound is produced in a big hall, its wave reflects from the walls and travel back and forth. Due to this, energy does not reduce and the sound persists. Small amount of reverberation for lesser time helps in adding volume to the programmers. Too much reverberation confuses the programmers and must be reduced.
To reduce reverberation, the roof and walls of the hall are covered with a sound absorbing material like rough plaster and thick curtains.
Q. Although the light is reflected from the book you read, why is your image not visible in it?
Ans: The rough surface diffuses or scatters the light falling on it and prevent the formation of image. Light is reflected from the paper of this book also but the surface of paper is much rougher than mirrors. That is why no image is formed by the paper.
Q. What is echo ranging? State any one application of this technique.
Echo ranging is the process of detecting underwater objects using sound signals. The minimum distance between source and the reflecting body should be 17 metres for the formation of an echo. This technique is used to measure depth of sea with the help of Sonar.
Q. A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compression from the source ?
Ans. Here, f = 500 Hz
The time interval between successive compression means the time period.
T =1/f =1/500 = 0.002 s
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 7
Solution: Area of the valley, A = 97280 km2
Level in the rise of water in the valley, h = 10 cm = (10/100000) km = (1/10000) km
Thus, amount of rain fall in 14 day = Ah = 97280 km2 × (1/10000) km = 9.828 km3
Amount of rain fall in 1 day = 9.828 /14 = 0.702km3
Volume of water in 3 rivers = length × breadth × height = 1072 km × 75 m × 3 m = 1072 km × (75/1000) km × (3/1000) km = 0.2412 km3 x3 = 0.7236 km3
This shows that the amount of rain fall is approximately equal to the amount of water in three rivers.
Optional Exercise – 13.5 - 10th Mathematics –Surface area and Volume
Q) The numerator of a fraction is one more than its denominator. If its reciprocal is subtracted from it,the difference is 11/30. Find the fraction.
Q) The numerator of a fraction is one more than its denominator. If its reciprocal is subtracted from it, the difference is 11/30. Find the fraction.
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