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MathsMediumMCQ2024 Β· 09 Apr Shift 1

Q62.If the sum of the series 1 + 1 + … + 1 is equal to 5 , then 50 d is equal to : 1β‹…(1+d) (1+d)(1+2 d) (1+9 d)(1+10 d) (1) 10 (2) 5 (3) 15 (4) 20

What This Question Tests

This question tests the ability to sum a series by recognizing that each term can be expressed as a difference of two terms (telescoping sum) using partial fraction decomposition.

Concepts Tested

Series sumPartial fraction decompositionTelescoping series

Formulas Used

1/(a(a+d)) = (1/d) * (1/a - 1/(a+d))

πŸ“š NCERT Sections This Tests

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5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )βˆ†xAnswer The initial kinetic energy of the bullet βˆ† x β†’ 0 βˆ‘ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1Γ—1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where β€˜lim’ stands for the limit of the sum when 2 βˆ†x tends to zero. Thus, for a varying force 2 Γ— 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳