A body of mass 1mathrmkg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions mathbfT_1 and mathbfT_2 , respectively, are (in N):
Suspended mass equilibrium with two angled strings
The diagram shows a suspended mass of 1 kg held by two strings making angles of 60 and 30 degrees with the horizontal.

Solution & Explanation

### Related Formula For a system in static equilibrium: sum F_x = 0 quad textand quad sum F_y = 0 ### Core Logic Let's resolve the tension forces vecT_1 and vecT_2 into horizontal and vertical components: - T_1 makes 60^circ with the horizontal. - T_2 makes 30^circ with the horizontal. - Downward gravitational force: W = m g = 1 times 10 = 10 \ mathrmN. 1. **Horizontal Equilibrium (sum F_x = 0):** T_1 cos 60^circ = T_2 cos 30^circ T_1 cdot frac12 = T_2 cdot fracsqrt32 implies T_1 = T_2 sqrt3 2. **Vertical Equilibrium (sum F_y = 0):** T_1 sin 60^circ + T_2 sin 30^circ = m g = 10 T_1 cdot fracsqrt32 + T_2 cdot frac12 = 10 ### Step 1: Solve for Tensions Substitute T_1 = T_2 sqrt3 into the vertical equilibrium equation: (T_2 sqrt3) fracsqrt32 + fracT_22 = 10 frac3 T_22 + fracT_22 = 10 implies 2 T_2 = 10 implies T_2 = 5 \ mathrmN Substitute T_2 back to obtain T_1: T_1 = 5 sqrt3 \ mathrmN Thus, the tension magnitudes are T_1 = 5sqrt3 \ mathrmN and T_2 = 5 \ mathrmN. ### Pattern Recognition Sees: Suspending particle static equilibrium with asymmetric strings. Trap: Associating components with incorrect trigonometry axes or swapping T_1 and T_2 in options. Shortcut: Since the incline of T_1 (60^circ) is steeper than that of T_2 (30^circ), T_1 must carry a larger portion of the load, meaning T_1 > T_2. From the choices, only (2) satisfies this hierarchy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion
Free body diagram showing force resolution of suspended mass
The diagram shows a suspended mass of 1 kg held by two strings making angles of 60 and 30 degrees with the horizontal.

Reference Study Guides

Laws Of Motion

More Laws of Motion Previous-Year Questions — Page 4

Q36 jee_main_2024_27_jan_morning Banking of Tracks
A train is moving with a speed of 12text m/s on rails which are 1.5text m apart. To negotiate a curve of radius 400text m, the height by which the outer rail should be raised with respect to the inner rail is (Given, g = 10text m/s^2):
  • A. 6.0text cm
  • B. 5.4text cm
  • C. 4.8text cm
  • D. 4.2text cm

Solution

### Related Formula tantheta = fracv^2Rg For small angles, tantheta approx sintheta = frachd, where h is the raised height and d is the separation between the tracks. ### Core Logic Equating the two relationships: frachd = fracv^2Rg frach1.5 = frac12 times 12400 times 10 ### Step 1: Compute height value h = 1.5 times frac1444000 = 1.5 times 0.036 = 0.054text m Converting to centimeters: h = 0.054 times 100 = 5.4text cm ### Pattern Recognition Whenever theta is small, geometry permits approximating tantheta with frachtextwidth, vastly reducing computational transcendental overhead. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion
Q45 jee_main_2024_27_jan_morning Conservation of Linear Momentum
A body of mass 1000text kg is moving horizontally with a velocity 6text m/s. If 200text kg extra mass is added, the final velocity (in m/s) is:
  • A. 6
  • B. 2
  • C. 3
  • D. 5

Solution

### Related Formula m_1 v_1 = m_2 v_2 ### Core Logic Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved. Initial mass m_1 = 1000text kg, initial velocity v_1 = 6text m/s. Final mass m_2 = 1000 + 200 = 1200text kg. ### Step 1: Balance conservation equation 1000 times 6 = 1200 times v_2 6000 = 1200 v_2 implies v_2 = frac60001200 = 5text m/s ### Pattern Recognition Inelastic mass addition transitions are classic momentum-balance equations, scaling velocity inversely with total expanded mass profiles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion
Q39 jee_main_2024_29_jan_morning Friction and Work Done
A block of mass 100 mathrm~kg slides over a distance of 10 mathrm~m on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction (in J) is:
  • A. 4200
  • B. 3900
  • C. 4000
  • D. 4500

Solution

### Related Formula Kinetic frictional force (f) on a flat surface: f = mu N = mu m g Work done against friction (W_textagainst): W_textagainst = f cdot s = mu m g s ### Core Logic Given values: m = 100 mathrm~kg, quad s = 10 mathrm~m, quad mu = 0.4 Take g = 10 mathrm~m/s^2. ### Step 1: Calculate Friction Force The normal force on a horizontal surface is: N = mg = 100 times 10 = 1000 mathrm~N The force of friction is: f = mu N = 0.4 times 1000 = 400 mathrm~N ### Step 2: Calculate Work Done The work done against friction is: W_textagainst = f cdot s = 400 mathrm~N times 10 mathrm~m = 4000 mathrm~J Therefore, the work done is 4000 mathrm~J. ### Pattern Recognition Work done *by* friction is negative (-4000 mathrm~J) because the frictional force opposes displacement. Work done *against* friction is positive (+4000 mathrm~J) because it represents the external energy that must be spent to sustain slide. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion
Q32 jee_main_2024_30_january_evening Work Done by Friction on Incline
A block of mass 1 mathrm~kg is pushed up a surface inclined to horizontal at an angle of 60^circ by a force of 10 mathrm~N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 mathrm~m along inclined surface, the work done against frictional force is: left[mathrmg = 10 mathrm~m / mathrms^2right]
Work Done by Friction on Incline diagram for Q32 - JEE Main 2024 Evening
A block of mass M on an incline at 60 degrees, pulled by 10 N force, with coefficient of static friction 0.1.
  • A. 5sqrt3 mathrm~J
  • B. 5 mathrm~J
  • C. 5 times 10^3 mathrm~J
  • D. 10 mathrm~J

Solution

### Related Formula W_f = f_k cdot d f_k = mu N N = mg costheta ### Core Logic The work done against the frictional force is the product of the kinetic friction force and the displacement along the plane. The normal force N on the block is given by N = mg costheta, where theta = 60^circ. ### Step 1: Calculate Normal and Frictional Force Given: mu = 0.1 m = 1 mathrm~kg theta = 60^circ d = 10 mathrm~m Normal force N = 1 times 10 times cos(60^circ) = 10 times frac12 = 5 mathrm~N. Frictional force f_k = mu N = 0.1 times 5 = 0.5 mathrm~N. ### Step 2: Work Done Against Friction Work done against frictional force = f_k times d W = 0.5 mathrm~N times 10 mathrm~m = 5 mathrm~J ### Pattern Recognition Whenever asked for "work done against friction", simply compute mu mg costheta times d. The applied force (10 mathrm~N) is irrelevant to the friction calculation itself since it is parallel to the plane. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Work, Energy and Power
Q39 jee_main_2024_30_january_evening Equilibrium on a Rough Parabolic Curve
A block of mass m is placed on a surface having vertical cross section given by y = x^2 / 4. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:
  • A. 1 / 4 mathrm~m
  • B. 1 / 2 mathrm~m
  • C. 1 / 6 mathrm~m
  • D. 1 / 3 mathrm~m

Solution

### Related Formula tan theta = mu fracmathrmdymathrmdx = tan theta ### Core Logic For a block to remain stationary on a rough surface without slipping, the maximum slope of the surface it can rest on is determined by the angle of repose. tan theta le mu The slope of the given parabolic curve at any point (x,y) is fracmathrmdymathrmdx. ### Step 1: Evaluate the Slope Given curve equation: y = fracx^24 Differentiating with respect to x: fracmathrmdymathrmdx = frac2x4 = fracx2 ### Step 2: Find Maximum Valid Position At the critical point of slipping, slope equals mu: fracx2 = mu Given mu = 0.5 = frac12, we have: fracx2 = frac12 implies x = 1 ### Step 3: Calculate Maximum Height Substitute x = 1 back into the curve equation to find the maximum height y: y = frac(1)^24 = frac14 mathrm~m ### Pattern Recognition When asked for maximum height on a curve y=f(x) without slipping, set fracmathrmdymathrmdx = mu, solve for x, and plug it back into the original equation to find y. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

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