371859
Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is
1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Explanation:
B Given, Forces \(5 \mathrm{~N}, 12 \mathrm{~N}, 13 \mathrm{~N}\) are in Equilibrium. Given, \(\sin 23^{\circ}=\frac{5}{13}\) We know, in \(\triangle \mathrm{ABC}\) \(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C} =180^{\circ}\) \(\alpha+90^{\circ}+23^{\circ} =180^{\circ}\) \(\alpha =180^{\circ}-113^{\circ}\) \(\alpha =67^{\circ}\)
Assam CEE-2014
LAWS OF MOTION (ADDITIONAL)
371860
The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are
1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)
Explanation:
A Let a and \(b\) is two forces. Then given, \(|\vec{a}|+|\vec{b}|=16 N,|\vec{R}|=8 N\) \(\therefore \quad \vec{R}=\vec{a}+\vec{b}\) Squaring both side, \((\vec{R})^{2}=(\vec{a}+\vec{b})^{2}\) \(|\overrightarrow{\mathrm{R}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \cos \theta}\) \(\text { Let, } \quad \vec{a}>\vec{b}\) \(\overrightarrow{\mathrm{R}} \cdot \overrightarrow{\mathrm{b}}=0\) \((\vec{a}+\vec{b}) \cdot \vec{b}=0\) \(\vec{a} \cdot \vec{b}=-(b)^{2}\) \(\mathrm{ab} \cos \square=-\mathrm{b}^{2}\) \(8=\sqrt{(16-b)^{2}+b^{2}+2(-b)^{2}}\) Squaring both side- \(64 =256+b^{2}-32 b+b^{2}-2 b^{2}\) \(32 b =256-64\) \(32 b =192\) \(b =6 \mathrm{~N}\) \(\therefore \quad a=16-6=10 \mathrm{~N}\)
AP EAMCET -2012
LAWS OF MOTION (ADDITIONAL)
371861
As shown in figure, the tension in the horizontal cord is \(30 \mathrm{~N}\). The weight \(W\) and tension in the string \(O A\) in Newton are
1 \(30 \sqrt{3}, 30\)
2 \(30 \sqrt{3}, 60\)
3 \(60 \sqrt{3}, 30\)
4 None of the above
Explanation:
B According to free body diagram - According to figure, \(\mathrm{W}=\mathrm{T} \cos 30^{\circ}\) \(30 =\mathrm{T} \sin 30^{\circ}\) \(\mathrm{T} =60 \mathrm{~N}\) Tension in cord, \(\mathrm{OA}=60 \mathrm{~N}\) \(\mathrm{W} =\mathrm{T} \cos 30^{\circ}\) \(\mathrm{W} =60 \times \frac{\sqrt{3}}{2}=30 \sqrt{3} \mathrm{~N}\) Hence, the correct answer is option (b).
JCECE-2016
LAWS OF MOTION (ADDITIONAL)
371862
A weight \(\mathrm{mg}\) is suspended from the middle of a rope whose ends are at same level. If the rope is no longer horizontal. The minimum tension required to completely straighten the rope will be
1 \(\mathrm{mg}\)
2 \(\sqrt{\mathrm{mg}}\)
3 Infinite
4 Zero
Explanation:
C According to free body diagram - According to figure, \(2 \mathrm{~T} \sin \theta=\mathrm{mg}\) \(\mathrm{T}=\frac{\mathrm{mg}}{2 \sin \theta}\) When rope is straight, \(\theta=0^{\circ}\) \(\because \quad \mathrm{T}=\frac{\mathrm{mg}}{2 \sin 0^{\circ}}=\infty\) This denotes that the minimum tension required to straighten the rope with weight suspended is infinite.
371859
Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is
1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Explanation:
B Given, Forces \(5 \mathrm{~N}, 12 \mathrm{~N}, 13 \mathrm{~N}\) are in Equilibrium. Given, \(\sin 23^{\circ}=\frac{5}{13}\) We know, in \(\triangle \mathrm{ABC}\) \(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C} =180^{\circ}\) \(\alpha+90^{\circ}+23^{\circ} =180^{\circ}\) \(\alpha =180^{\circ}-113^{\circ}\) \(\alpha =67^{\circ}\)
Assam CEE-2014
LAWS OF MOTION (ADDITIONAL)
371860
The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are
1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)
Explanation:
A Let a and \(b\) is two forces. Then given, \(|\vec{a}|+|\vec{b}|=16 N,|\vec{R}|=8 N\) \(\therefore \quad \vec{R}=\vec{a}+\vec{b}\) Squaring both side, \((\vec{R})^{2}=(\vec{a}+\vec{b})^{2}\) \(|\overrightarrow{\mathrm{R}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \cos \theta}\) \(\text { Let, } \quad \vec{a}>\vec{b}\) \(\overrightarrow{\mathrm{R}} \cdot \overrightarrow{\mathrm{b}}=0\) \((\vec{a}+\vec{b}) \cdot \vec{b}=0\) \(\vec{a} \cdot \vec{b}=-(b)^{2}\) \(\mathrm{ab} \cos \square=-\mathrm{b}^{2}\) \(8=\sqrt{(16-b)^{2}+b^{2}+2(-b)^{2}}\) Squaring both side- \(64 =256+b^{2}-32 b+b^{2}-2 b^{2}\) \(32 b =256-64\) \(32 b =192\) \(b =6 \mathrm{~N}\) \(\therefore \quad a=16-6=10 \mathrm{~N}\)
AP EAMCET -2012
LAWS OF MOTION (ADDITIONAL)
371861
As shown in figure, the tension in the horizontal cord is \(30 \mathrm{~N}\). The weight \(W\) and tension in the string \(O A\) in Newton are
1 \(30 \sqrt{3}, 30\)
2 \(30 \sqrt{3}, 60\)
3 \(60 \sqrt{3}, 30\)
4 None of the above
Explanation:
B According to free body diagram - According to figure, \(\mathrm{W}=\mathrm{T} \cos 30^{\circ}\) \(30 =\mathrm{T} \sin 30^{\circ}\) \(\mathrm{T} =60 \mathrm{~N}\) Tension in cord, \(\mathrm{OA}=60 \mathrm{~N}\) \(\mathrm{W} =\mathrm{T} \cos 30^{\circ}\) \(\mathrm{W} =60 \times \frac{\sqrt{3}}{2}=30 \sqrt{3} \mathrm{~N}\) Hence, the correct answer is option (b).
JCECE-2016
LAWS OF MOTION (ADDITIONAL)
371862
A weight \(\mathrm{mg}\) is suspended from the middle of a rope whose ends are at same level. If the rope is no longer horizontal. The minimum tension required to completely straighten the rope will be
1 \(\mathrm{mg}\)
2 \(\sqrt{\mathrm{mg}}\)
3 Infinite
4 Zero
Explanation:
C According to free body diagram - According to figure, \(2 \mathrm{~T} \sin \theta=\mathrm{mg}\) \(\mathrm{T}=\frac{\mathrm{mg}}{2 \sin \theta}\) When rope is straight, \(\theta=0^{\circ}\) \(\because \quad \mathrm{T}=\frac{\mathrm{mg}}{2 \sin 0^{\circ}}=\infty\) This denotes that the minimum tension required to straighten the rope with weight suspended is infinite.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
LAWS OF MOTION (ADDITIONAL)
371859
Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is
1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Explanation:
B Given, Forces \(5 \mathrm{~N}, 12 \mathrm{~N}, 13 \mathrm{~N}\) are in Equilibrium. Given, \(\sin 23^{\circ}=\frac{5}{13}\) We know, in \(\triangle \mathrm{ABC}\) \(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C} =180^{\circ}\) \(\alpha+90^{\circ}+23^{\circ} =180^{\circ}\) \(\alpha =180^{\circ}-113^{\circ}\) \(\alpha =67^{\circ}\)
Assam CEE-2014
LAWS OF MOTION (ADDITIONAL)
371860
The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are
1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)
Explanation:
A Let a and \(b\) is two forces. Then given, \(|\vec{a}|+|\vec{b}|=16 N,|\vec{R}|=8 N\) \(\therefore \quad \vec{R}=\vec{a}+\vec{b}\) Squaring both side, \((\vec{R})^{2}=(\vec{a}+\vec{b})^{2}\) \(|\overrightarrow{\mathrm{R}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \cos \theta}\) \(\text { Let, } \quad \vec{a}>\vec{b}\) \(\overrightarrow{\mathrm{R}} \cdot \overrightarrow{\mathrm{b}}=0\) \((\vec{a}+\vec{b}) \cdot \vec{b}=0\) \(\vec{a} \cdot \vec{b}=-(b)^{2}\) \(\mathrm{ab} \cos \square=-\mathrm{b}^{2}\) \(8=\sqrt{(16-b)^{2}+b^{2}+2(-b)^{2}}\) Squaring both side- \(64 =256+b^{2}-32 b+b^{2}-2 b^{2}\) \(32 b =256-64\) \(32 b =192\) \(b =6 \mathrm{~N}\) \(\therefore \quad a=16-6=10 \mathrm{~N}\)
AP EAMCET -2012
LAWS OF MOTION (ADDITIONAL)
371861
As shown in figure, the tension in the horizontal cord is \(30 \mathrm{~N}\). The weight \(W\) and tension in the string \(O A\) in Newton are
1 \(30 \sqrt{3}, 30\)
2 \(30 \sqrt{3}, 60\)
3 \(60 \sqrt{3}, 30\)
4 None of the above
Explanation:
B According to free body diagram - According to figure, \(\mathrm{W}=\mathrm{T} \cos 30^{\circ}\) \(30 =\mathrm{T} \sin 30^{\circ}\) \(\mathrm{T} =60 \mathrm{~N}\) Tension in cord, \(\mathrm{OA}=60 \mathrm{~N}\) \(\mathrm{W} =\mathrm{T} \cos 30^{\circ}\) \(\mathrm{W} =60 \times \frac{\sqrt{3}}{2}=30 \sqrt{3} \mathrm{~N}\) Hence, the correct answer is option (b).
JCECE-2016
LAWS OF MOTION (ADDITIONAL)
371862
A weight \(\mathrm{mg}\) is suspended from the middle of a rope whose ends are at same level. If the rope is no longer horizontal. The minimum tension required to completely straighten the rope will be
1 \(\mathrm{mg}\)
2 \(\sqrt{\mathrm{mg}}\)
3 Infinite
4 Zero
Explanation:
C According to free body diagram - According to figure, \(2 \mathrm{~T} \sin \theta=\mathrm{mg}\) \(\mathrm{T}=\frac{\mathrm{mg}}{2 \sin \theta}\) When rope is straight, \(\theta=0^{\circ}\) \(\because \quad \mathrm{T}=\frac{\mathrm{mg}}{2 \sin 0^{\circ}}=\infty\) This denotes that the minimum tension required to straighten the rope with weight suspended is infinite.
371859
Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is
1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Explanation:
B Given, Forces \(5 \mathrm{~N}, 12 \mathrm{~N}, 13 \mathrm{~N}\) are in Equilibrium. Given, \(\sin 23^{\circ}=\frac{5}{13}\) We know, in \(\triangle \mathrm{ABC}\) \(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C} =180^{\circ}\) \(\alpha+90^{\circ}+23^{\circ} =180^{\circ}\) \(\alpha =180^{\circ}-113^{\circ}\) \(\alpha =67^{\circ}\)
Assam CEE-2014
LAWS OF MOTION (ADDITIONAL)
371860
The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are
1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)
Explanation:
A Let a and \(b\) is two forces. Then given, \(|\vec{a}|+|\vec{b}|=16 N,|\vec{R}|=8 N\) \(\therefore \quad \vec{R}=\vec{a}+\vec{b}\) Squaring both side, \((\vec{R})^{2}=(\vec{a}+\vec{b})^{2}\) \(|\overrightarrow{\mathrm{R}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \cos \theta}\) \(\text { Let, } \quad \vec{a}>\vec{b}\) \(\overrightarrow{\mathrm{R}} \cdot \overrightarrow{\mathrm{b}}=0\) \((\vec{a}+\vec{b}) \cdot \vec{b}=0\) \(\vec{a} \cdot \vec{b}=-(b)^{2}\) \(\mathrm{ab} \cos \square=-\mathrm{b}^{2}\) \(8=\sqrt{(16-b)^{2}+b^{2}+2(-b)^{2}}\) Squaring both side- \(64 =256+b^{2}-32 b+b^{2}-2 b^{2}\) \(32 b =256-64\) \(32 b =192\) \(b =6 \mathrm{~N}\) \(\therefore \quad a=16-6=10 \mathrm{~N}\)
AP EAMCET -2012
LAWS OF MOTION (ADDITIONAL)
371861
As shown in figure, the tension in the horizontal cord is \(30 \mathrm{~N}\). The weight \(W\) and tension in the string \(O A\) in Newton are
1 \(30 \sqrt{3}, 30\)
2 \(30 \sqrt{3}, 60\)
3 \(60 \sqrt{3}, 30\)
4 None of the above
Explanation:
B According to free body diagram - According to figure, \(\mathrm{W}=\mathrm{T} \cos 30^{\circ}\) \(30 =\mathrm{T} \sin 30^{\circ}\) \(\mathrm{T} =60 \mathrm{~N}\) Tension in cord, \(\mathrm{OA}=60 \mathrm{~N}\) \(\mathrm{W} =\mathrm{T} \cos 30^{\circ}\) \(\mathrm{W} =60 \times \frac{\sqrt{3}}{2}=30 \sqrt{3} \mathrm{~N}\) Hence, the correct answer is option (b).
JCECE-2016
LAWS OF MOTION (ADDITIONAL)
371862
A weight \(\mathrm{mg}\) is suspended from the middle of a rope whose ends are at same level. If the rope is no longer horizontal. The minimum tension required to completely straighten the rope will be
1 \(\mathrm{mg}\)
2 \(\sqrt{\mathrm{mg}}\)
3 Infinite
4 Zero
Explanation:
C According to free body diagram - According to figure, \(2 \mathrm{~T} \sin \theta=\mathrm{mg}\) \(\mathrm{T}=\frac{\mathrm{mg}}{2 \sin \theta}\) When rope is straight, \(\theta=0^{\circ}\) \(\because \quad \mathrm{T}=\frac{\mathrm{mg}}{2 \sin 0^{\circ}}=\infty\) This denotes that the minimum tension required to straighten the rope with weight suspended is infinite.
