Reverted Gear Train MCQ Quiz in मराठी - Objective Question with Answer for Reverted Gear Train - मोफत PDF डाउनलोड करा
Last updated on Apr 21, 2025
Latest Reverted Gear Train MCQ Objective Questions
Top Reverted Gear Train MCQ Objective Questions
Reverted Gear Train Question 1:
In the given figure, if the speed of the input shaft of the spur gear train is 2400 rpm and the speed of the output shaft is 100 rpm, what is the module of the gear 4?
Answer (Detailed Solution Below)
Reverted Gear Train Question 1 Detailed Solution
The gear which meh must have the same circular pitch or module (D/T)
m1 = m2 = m = 1 mm
m3 = m4 = m’ =?
\(\begin{array}{l} \frac{{{D_1}}}{2} + \frac{{{D_2}}}{2} = 35\\ \frac{{m{T_1} + m{T_2}}}{2} = 35 \end{array}\)
m = 1, T1 = 60 ⇒ T2 = 10
\(\begin{array}{l} {N_1} = {N_2} \times \frac{{{T_2}}}{{{T_1}}} = 2400 \times \frac{{10}}{{60}} = 400 = {N_3}\\ \frac{{{N_3}}}{{{N_4}}} = \frac{{{T_4}}}{{{T_3}}} \Rightarrow {T_4} = \frac{{{N_3}}}{{{N_4}}} \times {T_3} = \frac{{400}}{{100}} \times 10 = 40\\ \frac{{{D_3}}}{2} + \frac{{{D_4}}}{2} = 35\\ \frac{{m'{T_3} + m'{T_4}}}{2} = 35\\ m'\left( {10 + 40} \right) = 70 \Rightarrow m' = 1.4 \end{array}\)
Reverted Gear Train Question 2:
In a reverted gear train the pinion (20 teeth) drives a gear B (60 teeth). The gear B and pinion C (15 teeth) are compounded. The pinion C drives the output gear D (45 teeth). The module for the first stage is 3. The module for the second stage and the velocity ratio are respectively:
Answer (Detailed Solution Below)
Reverted Gear Train Question 2 Detailed Solution
When the axes of the first gear (i.e. first driver) and the last gear (i.e. last driven) are co-axial, then the gear train is known as reverted gear train.
TA = 20, TB = 60, TC = 15, TD = 45
Distance between the centres of the shafts:
rA + rB = rC + rD
Module, \(m = \frac{{D\left( {mm} \right)}}{T}\)
m (TA + TB) = m’ (TC + TD)
3(20 + 60) = m’(15 + 45)
\(m' = \frac{{80 \times 3}}{{60}} = 4\)
Velocity Ratio:
\(\begin{array}{l} \frac{{{N_{Driver}}}}{{{N_{Driven}}}} = \frac{{Product\;of\;number\;of\;teeth\;on\;driven}}{{Product\;of\;number\;of\;teeth\;on\;drivers}}\\ V.R = \frac{{{N_A}}}{{{N_D}}} = \frac{{{T_B} \times {T_D}}}{{{T_A} \times {T_C}}} = \frac{{60 \times 45}}{{20 \times 15}} = 9 \end{array}\)
So, m’ = 4 and V.R = 9
Reverted Gear Train Question 3:
Two wheels A and B in the same straight line are geared together through an intermediate parallel shat. The parameters relating to the gears and pinions are given in the table
|
Item |
Speed |
Teeth |
PCD |
Module |
|
Driving wheel A |
NA |
TA |
DA |
m |
|
Driving wheel B |
NB |
TB |
DB |
m |
|
Driving wheel C on the intermediate shaft, in mesh with A |
NC |
TC |
DC |
m |
|
Driving wheel D on the intermediate shaft, in mesh with B |
ND |
TD |
DD |
m |
Which of the following relation is not valid for kinematic design consideration ?
Answer (Detailed Solution Below)
\(\frac{{{N_A}}}{{{N_B}}} = \frac{{{T_C}}}{{{T_A}}} \times \frac{{{T_B}}}{{{T_D}}}\)
Reverted Gear Train Question 3 Detailed Solution
Clearly
DA + DC = DB + DD
⇒ mTA + mTC = mTB + mTD
⇒ TA + TC = TB + TD
\(\frac{{{N_A}}}{{{N_B}}} = \frac{{{N_A}}}{{{N_C}}} \times \frac{{{N_C}}}{{{N_B}}} = \frac{{{N_A}}}{{{N_C}}} \times \frac{{{N_D}}}{{{N_B}}} = \frac{{{T_C}}}{{{T_A}}} \times \frac{{{T_B}}}{{{T_D}}}\)
Reverted Gear Train Question 4:
Gear A of gear train shown in figure is fixed. All the bevel gears are identical. If shaft- I, carrying the gear B on the arm as shown, rotates at 100 rpm in CCW direction. The speed and direction of rotation of shaft – II is
Answer (Detailed Solution Below)
200 rpm (CCW)
Reverted Gear Train Question 4 Detailed Solution
|
Step No. |
Conditions of motion |
Arm |
Gear A |
Gear B |
Gear C |
|
1 |
Arm is fixed, gear A rotated through + 1 revolution (i.e. 1 revolution anticlockwise) |
0 |
+1 |
\( \pm \frac{{{T_A}}}{{{T_B}}}\) |
\( - \frac{{{T_A}}}{{{T_B}}}.\frac{{{T_B}}}{{{T_C}}} = - \frac{{{T_A}}}{{{T_C}}}\) |
|
2 |
Arm fixed, gear A rotated through + x revolutions |
0 |
+x |
\( \pm x\frac{{{T_A}}}{{{T_B}}}\) |
\( -x \frac{{{T_A}}}{{{T_C}}}\) |
|
3 |
Add + y revolutions to all elements |
+y |
+y |
+y |
+y |
|
4 |
Total motion |
+y |
x+y |
\(y \pm x\frac{{{T_A}}}{{{T_B}}}\) |
\( y-x \frac{{{T_A}}}{{{T_C}}}\) |
Shaft – II and bevel gear C is a single unit.
And given that gears are identical. So, TA = TB = TC
Gear A is fixed: x+y=0
Arm rotates 100 rpm CCW; y=100
x=-100 (100 CW)
Speed of the driven shaft II i.e. Gear C:
\( y-x \frac{{{T_A}}}{{{T_C}}}=y-x=100-(-100)=200\; CCW\)
Reverted Gear Train Question 5:
For the figure given below, find the module of gear 4, If NInput= 2400 rpm and Noutput = 100 rpm. Given that module of gear-1 is 1 mm.
Answer (Detailed Solution Below)
1.4 mm
Reverted Gear Train Question 5 Detailed Solution
For the given gear train
r1 + r2 = r3 + r4 ……………….. (1)
and given that distance between shaft of gear-1 and input shaft is 70 mm (1, 2 are mating gears so module of 1, 2 is same).
So, r1 + r2 = 70 mm
\(\Rightarrow \frac{{m{T_1}}}{2} + \frac{{m{T_2}}}{2} = 70\)
\(\Rightarrow \frac{{1 \times 120}}{2} + \frac{{1 \times {T_2}}}{2} = 70 \Rightarrow {T_2} = 20\)
now, \(\frac{{{N_4}}}{{{N_2}}} = \frac{{{T_2}}}{{{T_1}}} \times \frac{{{T_3}}}{{{T_{4\;}}}} \Rightarrow \frac{{100}}{{2400}} = \frac{{200}}{{120}} \times \frac{{20}}{{{T_4}}} \Rightarrow {T_4} = 80\)
From equation (1), we get
r1 + r2 = r3 + r4 = 70
\(\Rightarrow \frac{{m'{T_3}}}{2} + \frac{{m'{T_4}}}{2} = 70\)
\(\Rightarrow \frac{{m'}}{2}\left( {20 + 80} \right) = 70 \Rightarrow m' = 1.4\)
since gear-3 and gear-4 are mating gear. Therefore their module will be same.
Hence, (C) is the correct choice.