The length of a light string is 1.4mathrmm when the tension on it is 5mathrmN . If the tension increases to 7mathrmN , the length of the string is 1.56mathrmm . The original length of the string is ______ mathrmm .

Numerical Answer Type:
Enter a numerical value Answer: 1 to 1 +4 marks

Solution & Explanation

### Related Formula By Hooke's Law, the tension (T) in a stretched elastic string is proportional to its change in length: T = k cdot Delta L = k(L - L_0) where: k = spring constant of the string L = current stretched length L_0 = natural original length ### Core Logic We can set up two algebraic equations using the given conditions: 1. **Case 1:** Tension T_1 = 5 \ mathrmN produces stretched length L_1 = 1.4 \ mathrmm: 5 = k(1.4 - L_0) quad dots (1) 2. **Case 2:** Tension T_2 = 7 \ mathrmN produces stretched length L_2 = 1.56 \ mathrmm: 7 = k(1.56 - L_0) quad dots (2) ### Step 1: Solve for natural length Divide equation (1) by equation (2) to eliminate the spring constant k: frac57 = frac1.4 - L_01.56 - L_0 Cross-multiply and solve for L_0: 5(1.56 - L_0) = 7(1.4 - L_0) 7.8 - 5 L_0 = 9.8 - 7 L_0 Rearrange to isolate the variable: 2 L_0 = 9.8 - 7.8 = 2 L_0 = 1 \ mathrmm Thus, the original length of the string is 1 mathrm~m. ### Pattern Recognition Sees: Elastic stretching of a string/wire under variable load. Trap: Attempting to resolve details like Young's modulus or cross-sectional area. Taking ratios allows the spring constant k to cancel cleanly, saving computational effort. Shortcut: Use the ratio equation fracT_1T_2 = fracL_1 - L_0L_2 - L_0 directly. Solving the linear equation yields L_0 = 1text m. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

Reference Study Guides

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Q58 jee_main_2024_31_jan_morning Bulk Modulus
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02\% is ________ mathrmm. (Take density of sea water = 10^3mathrm\ kgm^-3, Bulk modulus of rubber = 9 times 10^8mathrm\ Nm^-2, and g = 10mathrm\ ms^-2)
Numerical Answer. Answer: 18 to 18

Solution

### Related Formula beta = frac-Delta PfracDelta VV Delta P = rho g h ### Core Logic The change in pressure Delta P is the hydrostatic pressure at depth h. Delta P = -beta fracDelta VV rho g h = -beta fracDelta VV ### Step 2: Calculation Given values: rho = 10^3mathrm\,kg/m^3 g = 10mathrm\,m/s^2 beta = 9 times 10^8mathrm\,N/m^2 fracDelta VV = -0.02\% = -frac0.02100 Substitute into the equation: 10^3 times 10 times h = - (9 times 10^8) times left(-frac0.02100right) 10^4 times h = 9 times 10^8 times 2 times 10^-4 10^4 h = 18 times 10^4 h = 18mathrm\,m ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties Of Solids Class 11 Physics: Mechanical Properties Of Fluids

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