146007
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
146008
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.
JCECE-2013
Laws of Motion
145998
The magnitudes of a set of 3 vectors are given below. The set of vectors for which the resultant cannot be zero is
1 \(15,20,30\)
2 \(20,20,30\)
3 \(25,20,35\)
4 \(10,10,20\)
5 \(10,20,40\)
Explanation:
E According to the triangle law, the addition of first two vectors is greater than third vector. For the resultant cannot be zero. But in option (e) can not satisfied. i.e. \(\quad 10+20 \lt 40\)
Kerala CEE - 2015
Laws of Motion
146004
Which one among the following is the correct for resultant of balanced forces?
1 It is zero
2 It is non-zero
3 It varies continuously
4 None of the above
Explanation:
A Balanced forces- When two forces acting on an object are equal in size but act in opposite directions they are called balanced forces. \(\overrightarrow{\mathrm{F}}_{1} \stackrel{\mathrm{m}}{\longleftrightarrow} \stackrel{\rightharpoonup}{\mathrm{F}_{1}}-\overrightarrow{\mathrm{F}}_{2} \quad\left(\because \overrightarrow{\mathrm{F}}_{1}=\overrightarrow{\mathrm{F}}_{2}\right)\) \(\mathrm{F}_{\text {net }}=\overrightarrow{\mathrm{F}}_{1}-\overrightarrow{\mathrm{F}}_{1}\) \(\overrightarrow{\mathrm{F}}_{\text {net }}=0\) The resultant of balanced forces is zero.
146007
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
146008
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.
JCECE-2013
Laws of Motion
145998
The magnitudes of a set of 3 vectors are given below. The set of vectors for which the resultant cannot be zero is
1 \(15,20,30\)
2 \(20,20,30\)
3 \(25,20,35\)
4 \(10,10,20\)
5 \(10,20,40\)
Explanation:
E According to the triangle law, the addition of first two vectors is greater than third vector. For the resultant cannot be zero. But in option (e) can not satisfied. i.e. \(\quad 10+20 \lt 40\)
Kerala CEE - 2015
Laws of Motion
146004
Which one among the following is the correct for resultant of balanced forces?
1 It is zero
2 It is non-zero
3 It varies continuously
4 None of the above
Explanation:
A Balanced forces- When two forces acting on an object are equal in size but act in opposite directions they are called balanced forces. \(\overrightarrow{\mathrm{F}}_{1} \stackrel{\mathrm{m}}{\longleftrightarrow} \stackrel{\rightharpoonup}{\mathrm{F}_{1}}-\overrightarrow{\mathrm{F}}_{2} \quad\left(\because \overrightarrow{\mathrm{F}}_{1}=\overrightarrow{\mathrm{F}}_{2}\right)\) \(\mathrm{F}_{\text {net }}=\overrightarrow{\mathrm{F}}_{1}-\overrightarrow{\mathrm{F}}_{1}\) \(\overrightarrow{\mathrm{F}}_{\text {net }}=0\) The resultant of balanced forces is zero.
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Laws of Motion
146007
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
146008
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.
JCECE-2013
Laws of Motion
145998
The magnitudes of a set of 3 vectors are given below. The set of vectors for which the resultant cannot be zero is
1 \(15,20,30\)
2 \(20,20,30\)
3 \(25,20,35\)
4 \(10,10,20\)
5 \(10,20,40\)
Explanation:
E According to the triangle law, the addition of first two vectors is greater than third vector. For the resultant cannot be zero. But in option (e) can not satisfied. i.e. \(\quad 10+20 \lt 40\)
Kerala CEE - 2015
Laws of Motion
146004
Which one among the following is the correct for resultant of balanced forces?
1 It is zero
2 It is non-zero
3 It varies continuously
4 None of the above
Explanation:
A Balanced forces- When two forces acting on an object are equal in size but act in opposite directions they are called balanced forces. \(\overrightarrow{\mathrm{F}}_{1} \stackrel{\mathrm{m}}{\longleftrightarrow} \stackrel{\rightharpoonup}{\mathrm{F}_{1}}-\overrightarrow{\mathrm{F}}_{2} \quad\left(\because \overrightarrow{\mathrm{F}}_{1}=\overrightarrow{\mathrm{F}}_{2}\right)\) \(\mathrm{F}_{\text {net }}=\overrightarrow{\mathrm{F}}_{1}-\overrightarrow{\mathrm{F}}_{1}\) \(\overrightarrow{\mathrm{F}}_{\text {net }}=0\) The resultant of balanced forces is zero.
146007
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
146008
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.
JCECE-2013
Laws of Motion
145998
The magnitudes of a set of 3 vectors are given below. The set of vectors for which the resultant cannot be zero is
1 \(15,20,30\)
2 \(20,20,30\)
3 \(25,20,35\)
4 \(10,10,20\)
5 \(10,20,40\)
Explanation:
E According to the triangle law, the addition of first two vectors is greater than third vector. For the resultant cannot be zero. But in option (e) can not satisfied. i.e. \(\quad 10+20 \lt 40\)
Kerala CEE - 2015
Laws of Motion
146004
Which one among the following is the correct for resultant of balanced forces?
1 It is zero
2 It is non-zero
3 It varies continuously
4 None of the above
Explanation:
A Balanced forces- When two forces acting on an object are equal in size but act in opposite directions they are called balanced forces. \(\overrightarrow{\mathrm{F}}_{1} \stackrel{\mathrm{m}}{\longleftrightarrow} \stackrel{\rightharpoonup}{\mathrm{F}_{1}}-\overrightarrow{\mathrm{F}}_{2} \quad\left(\because \overrightarrow{\mathrm{F}}_{1}=\overrightarrow{\mathrm{F}}_{2}\right)\) \(\mathrm{F}_{\text {net }}=\overrightarrow{\mathrm{F}}_{1}-\overrightarrow{\mathrm{F}}_{1}\) \(\overrightarrow{\mathrm{F}}_{\text {net }}=0\) The resultant of balanced forces is zero.
