00:04Question says, at what temperature velocity of sound is double than that of at zero degree
00:13Celsius, a. 819 Kelvin, b. 819 degree Celsius, c. 600 degree Centigrade, d. 600 Kelvin. To calculate the speed of
00:38sound we will use
00:40formula, V is equal to under root gamma RT divided by molecular weight or mass. As we know that, speed
00:54of sound is
00:57directly proportional to the square root of temperature. At zero degree C, the speed V1 is equal to
01:09V and its other temperatures, its speed V2 will become double of its value. So, we have to find that
01:28temperature.
01:30Take a suppose, this is T1 which is obtained as 0 degree C which is equal to 273 Kelvin and
01:40we have to find T2.
01:45So, V1 upon V2 will be equal to under root T1 upon under root T2. As the value of V1
02:00is V and the value of V2
02:03is 2V is equal to under root T1. The value of T1 is 273 divided by under root T2. So,
02:15to remove the under root
02:22V2, we will get cancelled with V2. So, V2 will get cancelled with V2. So, V2 will get cancelled with
02:41V2. So, it will be 1 upon 4 is equal to
02:46273 divided by T2. So, we will take T2 at the left side of the equal. Here it is dividing.
03:02On the other side of equal, it will multiply.
03:04So, T2 upon 4 will be equal to 273 into 4. So, T2 will be equal to 273 into 4.
03:20So, T2 will be equal to 273 into 4. It will be
03:281092 Kelvin. Here the answer is not given in Kelvin. So, we will convert this into
03:40Celsius Celsius Celsius. So, T2 will be equal to 1092 minus 273 degree Celsius. So, T2 will be obtained as
03:56819 degree
04:00centigrade. So, this is our required answer. So, the correct option will be B, 819 degree centigrade.
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