Question
List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength = 0.29 m enters these structures at the point S and a sound detector is placed at D. Between the points S and D, the sound travels only through the tubes. List-II contains the possible smallest values of l (refer to the figures) for which the detector D records maximum amplitude. Ignore effects of sharp corners. [Given (15^ ) = 0.97] Choose the option that best describes the match between the entries in List-I to those in List-II. List-I List-II (P) (1) 1.32 m (Q) (2) 1.19 m (R) (3) 0.51 m (S) (4) 0.29 m (5) 0.13 m
Step-by-step solution
For maximum amplitude at the detector D, the path difference x between the two paths must be an integer multiple of the wavelength . For the smallest value of l, we take x = = 0.29 m. For configuration (P): The sound travels through a straight tube of length l and a semi-circular tube of diameter l. Path difference x = l 2 - l = l ( 2 - 1 ). Equating to : l ( 3.1416 2 - 1 ) = 0.29 l(0.5708) = 0.29 l 0.51 m. Thus, P 3. For configuration (Q): The sound travels through a straight tube of length l and a rectangular path of length 0.5l + l + 0.5l = 2l. Path difference x = 2l - l = l. Equating to : l = 0.29 m. Thus, Q 4. For configuration (R): The sound travels through a straight horizontal tube of length l and a path consisting of a vertical tube of length l followed by a semi-circular tube. The diameter of the semi-circle is the hypotenuse of the right triangle formed by the tubes, which is l^2 + l^2 = l 2 . The length of the semi-circular part is (l 2 ) 2 = l 2 . Path difference x = (l + l 2 ) - l = l 2 . Equating to : l 2 = 0.29 l = 0.29 2 0.29 1.414 3.1416 0.13 m. Thus, R 5. For configuration (S): The sound travels through a straight tube of length l and a triangular path. The angles of the triangle are 45^ , 105^ , and 180^ - (45^ + 105^ ) = 30^ . Using the sine rule, the lengths of the other two sides are l 30^ 105^ and l 45^ 105^ . Path difference x = l 30^ + 45^ 105^ - l. Given 105^ = 15^ = 0.97. x = l ( 0.5 + 1/ 2 0.97 - 1 ) l ( 0.5 + 0.707 0.97 - 1 ) = l ( 1.207 0.97 - 1 ) 0.244 l. Equating to : 0.244 l = 0.29 l 1.19 m. Thus, S 2. Matching the results, we get P 3, Q 4, R 5, S 2. Answer: P 3, Q 4, R 5, S 2