1. Intensity proportional to square of amplitude. Hence amplitude ratio is square root of intensity ratio. Hence ratio = square root of 0.3/0.7 = 0.65: D
2. Sound cannot be polarized:D
3. Intensity = Power / surface area = 10/4pi25 = 0.032 :A
4. Draw possible answers (only 2 ) for smaller freq. Given L = 1.5~ . The next possible L = 1~ and 0.5~ where ~ is wavelength and L is length of tube. Equating f~=speed, fL/1.5=f*L. The result is f = 1.5 f* which means present value is 1.5 times the new freq. The new freq has 3 antinodes( 2 ends and middle of tube). No need to consider the next result which is 2 antinodes (at both ends) : B
5.Draw straight passing the origin that passes through X and Y respectively. The line that passes Y has a lower resistance than that which passes X: A
6. If all bulbs working the voltage across L or N is smaller (combined resistance is 0.5 R) that M (resistance R) . When N breaks, the voltage across L and M are equal (both resistance R) . The voltage across L is 0.33 V before and 0.5 V after N breaks. Hence using P = V2/R, the bulb L is brighter. For M before break is 0.67V but after is 0.5V ,hence dimmer . V is the voltage across battery. : D
