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QJEEM21Ph5919: JEE Main 2021 Physics - Class 11 JEEM21Ph5919 Solution: The resultant of these forces and is approximately ..........…

The resultant of these forces OP→,OQ→,OR→,OS→\overrightarrow {OP} ,\overrightarrow {OQ} ,\overrightarrow {OR} ,\overrightarrow {OS} and OT→\overrightarrow {OT} is approximately .......... N.

[Take 3=1.7\sqrt 3 = 1.7 , 2=1.4\sqrt 2 = 1.4 Given i^\widehat i and j^\widehat j unit vectors along x, y axis]

JEE Main 2021 (Online) 27th August Morning Shift Physics - Vector Algebra Question 21 English

Multiple Choice Options

  • A 9.25i^+5j^9.25\widehat i + 5\widehat j
  • B 3i^+15j^3\widehat i + 15\widehat j
  • C 2.5i^−14.5j^2.5\widehat i - 14.5\widehat j
  • D −1.5i^−15.5j^- 1.5\widehat i - 15.5\widehat j

Step-by-Step Verified Solution

JEE Main 2021 (Online) 27th August Morning Shift Physics - Vector Algebra Question 21 English Explanation
Fx→=(10×32+20(12)+20(12)−15(12)−15(32))i^\overrightarrow {{F_x}} = \left( {10 \times {{\sqrt 3 } \over 2} + 20\left( {{1 \over 2}} \right) + 20\left( {{1 \over {\sqrt 2 }}} \right) - 15\left( {{1 \over {\sqrt 2 }}} \right) - 15\left( {{{\sqrt 3 } \over 2}} \right)} \right)\widehat i

=9.25i^= 9.25\widehat i

Fy→=(15(12)+20(32)+10(12)−15(12)−20(12))j^\overrightarrow {{F_y}} = \left( {15\left( {{1 \over 2}} \right) + 20\left( {{{\sqrt 3 } \over 2}} \right) + 10\left( {{1 \over 2}} \right) - 15\left( {{1 \over {\sqrt 2 }}} \right) - 20\left( {{1 \over {\sqrt 2 }}} \right)} \right)\widehat j

=5j^= 5\widehat j

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